678 lines
38 KiB
TeX
678 lines
38 KiB
TeX
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\section{Local Theory}
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This chapter aims to provide the reader with some of the essential notions and tools for the local
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theory needed to form a comprehensive understanding of the concept of hermitian manifolds and
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in particular, Kähler manifolds. The main focus will be on complex vector spaces and hermitian
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forms on those spaces and also the tools linear algebra provides.
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Later, we are going to focus on the different tangent bundles of complex manifolds, which are
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collections of vector spaces that vary in a geometric way on the manifold. Therefore, most of the
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notions and tools we are going to introduce in this chapter will be translated into the global
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context later.
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The main goal of this chapter will be the definition of the local Hodge star operator \nolinebreak
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$\hodgestar$ and the definition of the local Lefschetz and dual Lefschetz operator $L$ and
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$\Lambda$. For this purpose, we will start by providing a brief overview of the topic of
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complexification of vector spaces. Afterwards, we will focus on euclidean and hermitian vector
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spaces. Finally, we will conclude with the definition of the operators mentioned above.
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The reader should be aware that this chapter is based on the similarly named chapter in Daniel
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Huybrechts' \emph{Complex Geometry} \cite{Huybrechts2004}, although the level of detail found there
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is not to be expected.
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\subsection{Complexification of vector spaces}\;
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To simplify the notation, we are going to assume the following setting for the remainder of this
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section.
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\begin{set}
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Let $V$ denote a real $n$-dimensional vector space. Also, assume that $V$ is an almost complex
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vector space, i.e. $V$ is equipped with an
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endomorphism $I: V\rightarrow V$ such that $I^2 = -\id$. This endomorphism is called the almost
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complex structure of $V$.
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\end{set}
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Using this almost complex structure, we can also think of $V$ as a complex vector space with the
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$\mathbb{C}$-module structure defined as $(a+ib)\cdot v = a v + b I(v)$ for all $a,b \in
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\mathbb{R}$. For this complex vector space, we will write $(V,I)$. With the product rule for the
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determinant, we can calculate
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\begin{align*}
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\det(I)^2 = \det(I^2) = \det(-\id) = (-1)^n.
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\end{align*}
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Since $\det(I)$ is real, we conclude that $n = 2m$ for some $m \in \mathbb{N}$.
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Furthermore, $V$ and $(V,I)$ are equal as sets and if $(v_1,\dots,v_d)$ is a complex basis of
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$(V,I)$, it is immediate that $(v_1,I(v_1),\dots,v_d,I(v_d))$ is a real basis of $V$. Therefore,
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their dimensions relate as
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\begin{align*}
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\dim_\mathbb{C} (V,I) = d = \frac{1}{2} \dim_\mathbb{R} V = \frac{1}{2}n = m.
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\end{align*}
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Additionally, as an almost complex vector space, $V$ is endowed with a natural orientation. This
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boils down to the fact that the real space $\mathbb{C}^m$ has a natural orientation given by the
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basis $(e_1, ie_1, \dots e_m, ie_m)$, with the $e_1,\dots,e_m$ being the standard basis vectors (cf.
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\cite[Corollary 1.2.3]{Huybrechts2004}).
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At the same time, it is possible to construct a different complex vector space using $V$.
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\begin{defn}
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The \emph{complexification} $V_\mathbb{C}$ of $V$ is defined as
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$V_\mathbb{C} := V \otimes_\mathbb{R} \mathbb{C}$.
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\end{defn}
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Let $\left(v_1, \dots, v_n\right)$ be a real basis of $V$. With the properties of the tensor
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product, it is $\left(v_1 \otimes 1, \dots , v_n \otimes 1\right)$ a complex basis of
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$V_\mathbb{C}$. This shows that there exists an inclusion $V \into V_\mathbb{C}$ and for the
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dimension of $V_\mathbb{C}$, we get
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\begin{align*}
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\dim_\mathbb{C} V_\mathbb{C} = n = \dim_\mathbb{R} V.
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\end{align*}
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Now, the almost complex structure $I$ can be linearly extended to an almost complex structure
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$I_\mathbb{C}$ on $V_\mathbb{C}$. This is defined as $I_\mathbb{C} (v\otimes 1 + w \otimes i) :=
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I(v) \otimes 1 + I(w)\otimes i$, and it is evident that this linear extension also has the property
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$I_\mathbb{C}^2 = -\id$. Thus, we also call this to be an almost complex structure.
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\begin{nota}
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Note that for a vector $v\otimes \lambda \in V_\mathbb{C}$, it is a common practice to sometimes
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omit the tensor product in the notation, just noting $\lambda v$ instead of $v \otimes \lambda$. If
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it is possible without confusion, we will also write $I$ instead of $I_\mathbb{C}$ for the complex
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extension of the almost complex structure.
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\end{nota}
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The following proposition shows how the two $\mathbb{C}$-module structures on $V_\mathbb{C}$,
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defined by the almost complex structure $I$ and by multiplication with $i$, compare to each other.
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\begin{prop}[Decomposition of $V_\mathbb{C}$ {\cite[Lemma 1.2.5]{Huybrechts2004}}]
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\label{loc-theory:lm:decomposition-of-vc}
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For the complexification $V_\mathbb{C}$ we have the decomposition
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$V_\mathbb{C} = V^{1,0} \oplus V^{0,1}$ with
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\begin{align*}
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V^{1,0} := \left\{v \in V_\mathbb{C} \mid I(v) = iv\right\} \enspace
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\text{and} \enspace V^{0,1} := \left\{v \in V_\mathbb{C} \mid I(v) =- iv\right\}.
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\end{align*}
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\end{prop}
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\begin{proof}We extend the proof of the stated lemma in \cite{Huybrechts2004}.
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Let $v \in V_\mathbb{C}$. It is $v = \frac{1}{2} (v- iI(v)) + \frac{1}{2} (v + iI(v))$. A simple
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calculation shows
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\begin{align*}
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I(v-iI(v)) = I(v) -I(iI(v)) = I(v) - iI^2(v) = I(v) + iv = i(-iI(v) + v)
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\end{align*}
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and thus $\frac{1}{2} (v -iI(v)) \in V^{1,0}$. With a similar calculation we obtain
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$\frac{1}{2} (v+iI(v)) \in V^{0,1}$. At the same time, it holds to be $V^{1,0} \cap V^{0,1} = \{0\}$.
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Thus, the inclusion $V^{1,0} \oplus V^{0,1} \into V_\mathbb{C}$ is injective and with the above
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calculation, it is also surjective. Hence, it is a canonical isomorphism, so the decomposition is
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proven.
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\end{proof}
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\begin{rem} We expand the argument in the proof of \cite[Lemma 1.2.5]{Huybrechts2004}.
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The proof of the last proposition shows that a vector $w \in V^{1,0}$ can be written as
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$w = v-iI(v)$ for some $v \in V_\mathbb{C}$. At the same time, we can split $v = x + iy$
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with $x,y \in V$. Then it is
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\begin{align*}
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\overline{w} = \overline{v -iI_\mathbb{C}(v)} &= \overline{x + iy - i (I(x) + iI(y))}\\
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&= x -iy+iI(x)+I(y) = \overline{v}+ i(I(x) + -iI(y)) = \overline{v} + iI_\mathbb{C}(\overline{v}).
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\end{align*}
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Hence it is $\overline{w} \in V^{0,1}$. Similar calculations show that for $w \in V^{0,1}$, it is
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$\overline{w} \in V^{1,0}$ and $\overline{\widebar{w}} = w$. Since complex conjugation is
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$\mathbb{R}$-linear, this already proves that $V^{1,0}$ and $V^{0,1}$ are isomorphic as real vector
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spaces.
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\end{rem}
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\begin{rem}
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\label{loc-theory:rem:c-linear-c-antilinear}
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Using the proof of the last proposition and the natural inclusion \linebreak$V \into V_\mathbb{C},\,
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v \mapsto v \otimes 1$, we can define an $\mathbb{R}$-linear isomorphism
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\begin{align*}
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\varphi_1: (V,I) \rightarrow V^{1,0},\;\enspace v \mapsto \big(v\otimes 1-iI_\mathbb{C}(v\otimes 1)\big).
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\end{align*}
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However, we are able to calculate
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\begin{align*}
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\varphi_1(I(v)) = I(v) \otimes 1 - iI_\mathbb{C}(I(v) \otimes 1) = I_\mathbb{C}(v \otimes 1)
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-iI_\mathbb{C}^2(v \otimes 1) = I_\mathbb{C}\big(v\otimes 1 -iI_\mathbb{C}(v\otimes 1)\big).
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\end{align*}
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Hence, we obtain $\varphi_1(I(v)) = I_\mathbb{C}(\varphi_1(v)) = i\varphi_1(v)$ because
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$\varphi_1(v) \in V^{1,0}$. Since the $\mathbb{C}$-module structure on $(V,I)$ is defined using $I$,
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we know that $\varphi_1$ is also a $\mathbb{C}$-linear isomorphism. At the same time, we are able to
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define a similar $\mathbb{R}$-linear isomorphism
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\begin{align*}
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\varphi_2: (V,I) \rightarrow V^{0,1} \;\enspace
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v \mapsto (v \otimes 1 + iI_\mathbb{C}(v \otimes 1)).
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\end{align*}
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The same calculation yields $\varphi_2(I(v)) = -i\varphi_2(v)$. Thus, $\varphi_2$ is a
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$\mathbb{C}$-antilinear isomorphism.
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\end{rem}
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Next, we are going to define an induced almost complex structure on the dual space $V^*$. Because of
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its induced nature, this almost complex structure is also called $I$ and it is defined as a mapping
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$I: V^* \rightarrow V^*$, such that $I(f)(v) = f(I(v))$ for all $f \in V^*$ and $v \in V$.
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Now, the following lemma ensures the compatibility of the complexification with the dual space of $V$.
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\begin{lm}[{\cite[Lemma 1.2.6]{Huybrechts2004}}]
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\label{loc-theory:lm:compatibility-of-dual-and-complexification}
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It is $(V_\mathbb{C})^* = \Hom_\mathbb{R}(V,\mathbb{C}) = (V^*)_\mathbb{C}$ and it also holds to be
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\begin{alignat*}{2}
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(V^{1,0})^* &= \left\{f \in \Hom_\mathbb{R}(V,\mathbb{C}) \mid f(I(v)) = if(v)\; \forall v \in V\right\} = (V^*)^{1,0},\\
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(V^{0,1})^* &= \left\{f \in \Hom_\mathbb{R}(V,\mathbb{C}) \mid f(I(v)) = -if(v)\; \forall v \in V\right\} = (V^*)^{0,1}.
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\end{alignat*}
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\end{lm}
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\begin{proof}
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It is $(V_\mathbb{C})^* = \Hom_\mathbb{C} (V_\mathbb{C}, \mathbb{C})$ and
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$(V^*)_\mathbb{C} = \Hom_\mathbb{R} (V, \mathbb{C})$. In order to prove that these two spaces
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are equal, we have to prove the existence of a canonical isomorphism
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$\Hom_\mathbb{R}(V,\mathbb{C}) \cong \Hom_\mathbb{C} (V_\mathbb{C}, \mathbb{C})$.
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Let $f \in \Hom_\mathbb{R}(V,\mathbb{C})$ and extend it to an $\mathbb{R}$-linear mapping
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$\tilde f: V_\mathbb{C} \rightarrow \mathbb{C}$ by setting $\tilde f(v \otimes \lambda): = \lambda
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f(v)$ for all $v \in V$ and $\lambda \in \mathbb{C}$. This mapping is also $\mathbb{C}$-linear
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because we can show for all $\mu \in \mathbb{C}$
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\begin{align*}
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\tilde f(\mu\cdot(v \otimes \lambda)) = \tilde f (v \otimes \mu\lambda) = \mu\lambda f(v) = \mu
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\cdot \tilde f (v \otimes \lambda).
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\end{align*}
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This shows that for every $f$, we can find a unique $\tilde f \in \Hom_\mathbb{C}(V_\mathbb{C},
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\mathbb{C})$.\\
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Let now $g \in \Hom_\mathbb{C}(V_\mathbb{C}, \mathbb{C})$. Using the inclusion
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$V \into V_\mathbb{C}$, we can restrict $g$ to obtain a mapping $h: V \rightarrow \mathbb{C}$
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that is defined as $h(v):= g(v \otimes 1)$.
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Since $g$ was $\mathbb{C}$-linear, $h$ is already an $\mathbb{R}$-linear mapping.
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This shows that $h \in \Hom_\mathbb{R}(V,\mathbb{C})$, and since those two constructions are
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obviously inverse to each other, this completes the proof of the first statement.\footnote{This
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first part of the proof was created using a Math Stack Exchange post of the user \emph{Mark}
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(\url{https://math.stackexchange.com/users/470733/mark}) that can be found on
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\url{https://math.stackexchange.com/q/4718935} and was last checked on the 25th of August, 2023.}\\
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For the second statement, we use \Cref{loc-theory:rem:c-linear-c-antilinear} to get
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\begin{align*}
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(V^{1,0})^* = \Hom_\mathbb{C}(V^{1,0},\mathbb{C})&= \Hom_\mathbb{C}((V,I),\mathbb{C}) \\&=
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\left\{f \in \Hom_\mathbb{R}(V,\mathbb{C}) \mid f(I(v)) = if(v)\; \forall v \in V\right\}.
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\end{align*}
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Additionally, for the other subspace, we can use the same remark to obtain
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\begin{align*}
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(V^{0,1})^* = \Hom_\mathbb{C}(V^{0,1},\mathbb{C}) &= \Hom_{\overline{\mathbb{C}}}((V,I),
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\mathbb{C}) \\&= \left\{f \in \Hom_\mathbb{R}(V,\mathbb{C})\mid f(I(v)) = -if(v)\; \forall v \in
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V\right\}.
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\end{align*}
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\end{proof}
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\begin{nota}
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Because of the last lemma, we will only write $V^*_\mathbb{C}$, omitting the brackets from now on.
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\end{nota}
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\subsection{Euclidian and hermitian vector spaces}\;
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Later, we are going to define the notion of a hermitian manifold, i.e. a complex manifold whose
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holomorphic tangent space in every point is equipped with a hermitian form. In order to do so, this
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section will cover some fundamental statements about those forms on complex vector spaces.
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For the remainder of this section, we are going to assume the following setting.
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\begin{set}
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Let $(V,g)$ be a real $n$-dimensional euclidean vector space, i.e. $g$ is a positive definite
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symmetric bilinear form on the real space $V$. Also, assume that $V$ is equipped with an almost
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complex structure $I$.
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\end{set}
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\begin{defn}
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The inner product $g$ is said to be \emph{compatible with the almost complex structure} $I$ if it
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holds to be $g(I(v),I(w)) = g(v,w)$ for all $v,w \in V$.
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\end{defn}
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\begin{nota}
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If the inner product $g$ on $V$ is compatible with the almost complex structure $I$, we usually
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only write $(V,g,I)$.
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\end{nota}
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The just-established notion of a compatible inner product gives rise to an additional notion.
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\begin{defn}
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The \emph{fundamental form} associated to $(V,g,I)$ is defined as the form $\omega \in
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\bigwedge^2V^* \cap \bigwedge^{1,1} V^*$, such that for all $v,w \in V$ it is
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\begin{align*}
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\omega(v,w) := g(I(v),w) = - g(v, I(w)).
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\end{align*}
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Note that the second equality is equivalent to $g$ being compatible with the almost complex
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structure $I$. Also, note that this immediately yields $\omega(I(v),I(w)) = \omega(v,w)$.
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\end{defn}
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\begin{rem}
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The expression $ \bigwedge^2V^* \cap \bigwedge^{1,1} V^*$ has to be explained. With
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\Cref{loc-theory:lm:compatibility-of-dual-and-complexification}, we know that
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$\bigwedge^2 V^*\subset \bigwedge^2 V_\mathbb{C}^*$. At the same time, it is
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\begin{align}
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\label{loc-theory:eq:decomps-of-complex-2-forms}
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\bigwedge\nolimits^2 V_\mathbb{C}^* = \bigwedge\nolimits^{2,0} V^* \oplus \bigwedge\nolimits^{1,1}
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V^* \oplus \bigwedge\nolimits^{0,2} V^*
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\end{align}
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(cf. \cite[Proposition 1.2.8 (ii), Example 1.2.34]{Huybrechts2004}).
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and for these reasons, the intersection is meaningful as it happens in $\bigwedge^2V^*_\mathbb{C}$.
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This expression describes all the alternating real 2-forms of type $(1,1)$, i.e. alternating real
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2-forms on $V$, that are also $\mathbb{C}$-linear in it's first argument and $\mathbb{C}$-antilinear
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in its second argument if viewed as forms on $V_\mathbb{C}$.
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\end{rem}
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Now, for the fundamental form $\omega$ from the last definition to be well-defined, we have
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to check whether $\omega$ is indeed an alternating real 2-form on $V$ and is also of type $(1,1)$.
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For the first statement, real bilinearity follows directly with the bilinearity of $g$. Also using the symmetry
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of $g$, we calculate for all $v,w \in V$
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\begin{align*}
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\omega(v,w) = g(I(v),w) = g(I^2(v),I(w)) = - g(v,I(w)) = - g(I(w),v) = -\omega(w,v).
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\end{align*}
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Hence, $\omega$ is alternating and therefore a real 2-form. With
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\Cref{loc-theory:eq:decomps-of-complex-2-forms}, it suffices to show that the $\mathbb{C}$-bilinear
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extension of $\omega$ vanishes on all pairs of vectors $v,w$ in $V^{1,0}$ or $V^{0,1}$ to
|
|||
|
|
prove that it is of type $(1,1)$. In the first case, i.e. $v,w \in V^{1,0}$, we calculate
|
|||
|
|
\begin{align*}
|
|||
|
|
\omega(v,w) = \omega(I(v),I(w)) = \omega(iv,iw) = i^2 \omega(v,w) = -\omega(v,w).
|
|||
|
|
\end{align*}
|
|||
|
|
Hence $\omega(v,w) = 0$. The first equation holds because the complex bilinear extension
|
|||
|
|
inherits this property from the real form $\omega$. The calculation for the other case can be carried out
|
|||
|
|
analogously. Hence, $\omega$ is indeed of type $(1,1)$ and this establishes the well-definedness of
|
|||
|
|
$\omega$.
|
|||
|
|
|
|||
|
|
For $(V,g,I)$, we can also define a positive definite hermitian form on the complex space $(V,I)$.
|
|||
|
|
This form is defined as
|
|||
|
|
\begin{align*}
|
|||
|
|
h: (V,I) \times (V,I) \rightarrow \mathbb{C},\;\enspace
|
|||
|
|
(v,w) \mapsto g(v,w) - i\omega(v,w).
|
|||
|
|
\end{align*}
|
|||
|
|
Additionally, the inner product $g$ on $V$ can be extended sesquilinearly to a positive definite
|
|||
|
|
hermitian form on $V_\mathbb{C}$. This extension is defined as
|
|||
|
|
\begin{align*}
|
|||
|
|
h_\mathbb{C} : V_\mathbb{C} \times V_\mathbb{C} \rightarrow \mathbb{C},\; \enspace
|
|||
|
|
(v \otimes \lambda, w \otimes \mu) \mapsto (\lambda \overline{\mu}) \cdot g(v,w).
|
|||
|
|
\end{align*}
|
|||
|
|
See also \cite[p.\,30]{Huybrechts2004} for the similar definitions.
|
|||
|
|
However, it has to be checked whether these two positive definite hermitian forms are well-defined.
|
|||
|
|
\begin{prop}[{\cite[Lemma 1.2.15]{Huybrechts2004}}]
|
|||
|
|
For $(V,g,I)$, the form $h:= g- i\omega$ is indeed a positive definite
|
|||
|
|
hermitian form on $(V,I)$. Also, the extension $h_\mathbb{C}$ of the inner product $g$ defines a
|
|||
|
|
positive definite hermitian form on $V_\mathbb{C}$.
|
|||
|
|
\end{prop}
|
|||
|
|
\begin{proof}
|
|||
|
|
We expand the calculation in the proof of the stated lemma in \cite{Huybrechts2004} and add the
|
|||
|
|
argument for our second statement. Let
|
|||
|
|
$v,w \in (V,I)$. It is $h(v,v) = g(v,v) - i \omega(v,v) =g(v,v)$ because $\omega$ is alternating and
|
|||
|
|
therefore $\omega(v,v) = 0$. Since $g$ is positive definite, this proves that $h$ is positive definite
|
|||
|
|
as well. Furthermore, since $g$ is symmetric, it is also
|
|||
|
|
\begin{align*}
|
|||
|
|
h(v,w) = g(v,w) - i\omega(v,w) = g(w,v) + i\omega(w,v) = \overline{h(w,v)},
|
|||
|
|
\end{align*}
|
|||
|
|
and it also holds to be
|
|||
|
|
\begin{align*}
|
|||
|
|
h(I(v),w) &= g(I(v),w) - i\omega(I(v),w) \\&= g(I^2(v),I(w)) -i (g(I^2(v),w)) \\&= -g(v,I(w)) +
|
|||
|
|
ig(v,w) \\&= i\big(i g(v,I(w)) + g(v,w)\big) \\&= i\big(- i\omega(v,w) + g(v,w)\big) = i h(v,w).
|
|||
|
|
\end{align*}
|
|||
|
|
On $(V,I)$, the image under $I$ corresponds to multiplication with $i$ because the
|
|||
|
|
$\mathbb{C}$-module structure is defined using the almost complex structure. This proves the
|
|||
|
|
$\mathbb{C}$-linearity in the first argument, as the $\mathbb{R}$-linearity is already inherited from
|
|||
|
|
$g$ and $\omega$.
|
|||
|
|
|
|||
|
|
For the $\mathbb{C}$-antilinearity in the second argument, we combine the last two statements to
|
|||
|
|
get
|
|||
|
|
\begin{align*}
|
|||
|
|
h(v,I(w)) = \overline{h(I(w),v)} = \overline{ih(w,v)} = -ih(v,w).
|
|||
|
|
\end{align*} This completes the proof of the first statement.
|
|||
|
|
|
|||
|
|
To prove the second statement, it is already clear by definition that $h_\mathbb{C}$ is
|
|||
|
|
$\mathbb{C}$-linear in its first argument and $\mathbb{C}$-antilinear in its second argument. Let
|
|||
|
|
$(v_1,\dots,v_n)$ be an orthonormal basis of $V$ with respect to the inner product $g$. With the
|
|||
|
|
properties of the tensor product, it is again $(v_1 \otimes 1, \dots, v_n \otimes 1)$ a basis of
|
|||
|
|
$V_\mathbb{C}$. Therefore, we can write every element $u \in V_\mathbb{C}$ as
|
|||
|
|
$u = \sum_{j=1}^{n} \lambda_j(v_j \otimes 1)= \sum_{j=1}^{n} v_j \otimes \lambda_j$. We are then
|
|||
|
|
able to calculate
|
|||
|
|
\begin{align*}
|
|||
|
|
h_\mathbb{C}(u,u) = h_\mathbb{C}\Big(\sum_{j=1}^n v_j \otimes \lambda_j, \sum_{k=1}^n v_k \otimes
|
|||
|
|
\lambda_k\Big) = \sum_{j,k=1}^{n} \lambda_j\overline\lambda_k g(v_j,v_k) = \sum_{j=1}^n|\lambda_j|^2
|
|||
|
|
\geq 0,
|
|||
|
|
\end{align*}
|
|||
|
|
|
|||
|
|
and $h_\mathbb{C}(u,u) = 0$ if and only if $u = 0$. Hence, $h_\mathbb{C}$ is positive definite.
|
|||
|
|
Furthermore, it holds to be
|
|||
|
|
\begin{align*}
|
|||
|
|
h_\mathbb{C}(v \otimes \lambda,w \otimes \mu) = \lambda \overline{\mu} \cdot g(v,w) = \overline
|
|||
|
|
{\overline \lambda \mu \cdot g(v,w)} = \overline{\overline \lambda \mu \cdot g(w,v)} =
|
|||
|
|
\overline{h_\mathbb{C} (w \otimes \mu, v \otimes \lambda)}.
|
|||
|
|
\end{align*}
|
|||
|
|
Thus, $h$ and $h_\mathbb{C}$ are both positive definite hermitian forms.
|
|||
|
|
\end{proof}
|
|||
|
|
|
|||
|
|
\begin{nota} In \cite[Lemma 1.2.17]{Huybrechts2004}, it is shown that these two hermitian forms only
|
|||
|
|
differ by a factor of $\frac{1}{2}$ under the natural inclusion $(V,I) \into V^{1,0}$. This may be a
|
|||
|
|
reason for the common practice not to differentiate between $h$ and $h_\mathbb{C}$ in the notation,
|
|||
|
|
which we will also adhere to.
|
|||
|
|
\end{nota}
|
|||
|
|
\begin{rem}
|
|||
|
|
\label{loc-theory:rem:real-of-hermitian-form}
|
|||
|
|
In the last proposition, it has been proven that the compatible inner product $g$ already defines a
|
|||
|
|
positive definite hermitian form on $(V,I$). Let now $\tilde h$ be an arbitrary positive definite hermitian
|
|||
|
|
form on $(V,I)$. We can define the \emph{real part} of $h$ as follows.
|
|||
|
|
\begin{align*}
|
|||
|
|
\Re(\tilde h)(v,w) := \frac{1}{2} \big(\tilde h(v,w) + \overline{\tilde h(v,w)}\big) =
|
|||
|
|
\frac{1}{2}\big(\tilde h(v,w) + \tilde h(w,v)\big)
|
|||
|
|
\end{align*}
|
|||
|
|
With the second equality, it is obvious that this defines a real positive definite and symmetric
|
|||
|
|
bilinear form on $V$. It is also
|
|||
|
|
\begin{align*}
|
|||
|
|
\Re(\tilde h)(I(v),I(w)) &= \frac{1}{2} \big(\tilde h(I(v),I(w)) + \tilde h(I(w),I(v))\big)\\&=
|
|||
|
|
\frac{1}{2} \big(\tilde h(v,w) + \tilde h(w,v)\big) \\&= \Re(\tilde h)(v,w)
|
|||
|
|
\end{align*}
|
|||
|
|
and thus $\Re(\tilde h)$ defines a compatible inner product on $V$. Because these constructions are
|
|||
|
|
inverse to each other, there is a one-to-one relation between positive definite hermitian forms on
|
|||
|
|
$(V,I)$ and inner products on $V$ that are compatible with $I$. See also
|
|||
|
|
\cite[Section. 3.1.1]{Voisin2002} for further information about this relation.
|
|||
|
|
\end{rem}
|
|||
|
|
|
|||
|
|
Later, we will also need inner products and hermitian forms on the exterior algebra spaces
|
|||
|
|
$\bigwedge^kV^*$ and $\bigwedge^k V^*_\mathbb{C}$. In the following two lemmas, we are going
|
|||
|
|
to construct inner products on $V^*$ and $\bigwedge^kV$ using the existing inner product $g$ on $V$,
|
|||
|
|
and we will eventually use those constructions to define an induced inner product on
|
|||
|
|
$\bigwedge\nolimits^kV^*$. This construction is inspired by \cite[Section 11]{Schnell2012}, but no
|
|||
|
|
proofs are provided there.
|
|||
|
|
|
|||
|
|
Before we begin, we need to take a look at the natural linear mapping
|
|||
|
|
\begin{align*}
|
|||
|
|
g^\flat: V \rightarrow V^*, \;v\mapsto g(v,-).
|
|||
|
|
\end{align*}
|
|||
|
|
With respect to the inner product $g$, we can choose an orthonormal basis $(v_1,\dots,v_n)$ of $V$.
|
|||
|
|
For all $r,s \in \mathbb{N}$ with $1 \leq r,s \leq n$, we get
|
|||
|
|
\begin{align*}
|
|||
|
|
g^\flat(v_r)(v_s) = g(v_r,v_s) = \delta_{rs}.
|
|||
|
|
\end{align*}
|
|||
|
|
Hence, $g^\flat(v_r) = v^r$ with $v^r$ being the dual basis vector of $v_r$. This already shows that
|
|||
|
|
$g^\flat$ is an isomorphism. Let now $g^\sharp: V^* \rightarrow V$ denote the inverse mapping of
|
|||
|
|
$g^\flat$. Using this mapping, we get the following lemma.
|
|||
|
|
\begin{lm}
|
|||
|
|
\label{loc-theory:lm:product-on-dual-space}
|
|||
|
|
The inner product $g$ on $V$ induces an inner product on $V^*$. It is defined \nolinebreak as
|
|||
|
|
\begin{align*}
|
|||
|
|
\tilde{g}: V^* \times V^* \rightarrow \mathbb{R},\; \enspace
|
|||
|
|
(v,w) \mapsto g\big(g^\sharp(v),g^\sharp(w)\big).
|
|||
|
|
\end{align*}
|
|||
|
|
\end{lm}
|
|||
|
|
\begin{proof}
|
|||
|
|
It is obvious that $\tilde{g}$ defines a bilinear mapping. Let now $(v_1,\dots,v_n)$ be an
|
|||
|
|
orthonormal basis of $V$ again. Also let $(v^1,\dots,v^n)$ denote the corresponding dual basis of
|
|||
|
|
$V^*$. If only evaluated on those dual basis vectors, $\tilde g$ simplifies as
|
|||
|
|
\begin{align*}
|
|||
|
|
\tilde{g}(v^r,v^s) = g(g^\sharp(v^r), g^\sharp(v^s)) = g(v_r,v_s).
|
|||
|
|
\end{align*}
|
|||
|
|
Thus, $\tilde{g}$ directly inherits the inner product properties from \nolinebreak$g$.
|
|||
|
|
\end{proof}
|
|||
|
|
\begin{lm}
|
|||
|
|
\label{loc-theory:lm:product-on-exterior-algebra}
|
|||
|
|
The inner product $g$ on $V$ induces an inner product on $\bigwedge\nolimits^kV$, which is defined as
|
|||
|
|
\begin{align*}
|
|||
|
|
g_k: \bigwedge\nolimits^kV \times \bigwedge\nolimits^kV &\rightarrow \mathbb R\\
|
|||
|
|
\big(v_1 \wedge \dots \wedge v_k, w_1\wedge\dots \wedge w_k\big) &\mapsto
|
|||
|
|
\det\Big(\big(g(v_r,w_s)\big)_{rs}\Big).
|
|||
|
|
\end{align*}
|
|||
|
|
\end{lm}
|
|||
|
|
\begin{proof}
|
|||
|
|
With the multilinearity of the determinant, it is again obvious that $g_k$ is a bilinear mapping.
|
|||
|
|
Since the determinant is invariant under transposition, $g_k$ is also symmetric. Let now
|
|||
|
|
$(v_1,\dots,v_n)$ be an orthonormal basis of $V$ again. We know that this induces a basis
|
|||
|
|
$(v_J)_J$ of $\bigwedge^k V$. (cf. \cite[Proposition 14.8]{Lee2012}.) If we only evaluate $g_k$
|
|||
|
|
on these basis vectors again, we get
|
|||
|
|
\begin{align*}
|
|||
|
|
g_k(v_J,v_K) = \det\Big(\big(g(v_{j_r}, v_{k_s})\big)_{rs}\Big) =
|
|||
|
|
\delta_{JK}.
|
|||
|
|
\end{align*}
|
|||
|
|
This is because if $J \neq K$, there is at least one $j_l \in J$ such that $j_l \not\in K$. Thus we
|
|||
|
|
obtain $g(v_{j_l}, v_{\tilde{k}}) = 0$ for all $\tilde{k} \in K$. Hence, the $l$-th column of the
|
|||
|
|
matrix $G:=\big(g(v_{j_r},v_{k_s})\big)_{rs}$ is everywhere zero and therefore
|
|||
|
|
$\det(G) = 0$. If $J$ and $K$ are equal however, then it is $G = \id_k$ and therefore
|
|||
|
|
$\det(G) = 1$. With this, we have for all $\alpha := \sum_{J} \alpha_J v_J\in \bigwedge^k V$
|
|||
|
|
\begin{align*}
|
|||
|
|
g_k(\alpha,\alpha) = \sum_{J} \alpha_J^2\, g_k(v_J,v_J) = \sum_{J} \alpha_J^2 \geq 0.
|
|||
|
|
\end{align*}
|
|||
|
|
This calculation also implies that $g_k(\alpha,\alpha) = 0$ if and only if $\alpha = 0$. Thus, the
|
|||
|
|
statement is proven.
|
|||
|
|
\end{proof}
|
|||
|
|
The combination of the last two lemmas finally gives us an inner product on
|
|||
|
|
$\bigwedge\nolimits^kV^*$.
|
|||
|
|
\begin{cor}
|
|||
|
|
\label{loc-theory:cor:induced-product-on-exterior-algebra}
|
|||
|
|
The induced inner product on $\bigwedge^kV^*$ is given as
|
|||
|
|
\begin{align*}
|
|||
|
|
\tilde{g}_k: \bigwedge^kV^* \times \bigwedge^kV^* &\rightarrow \mathbb{R}\\
|
|||
|
|
\big(v^1 \wedge \dots \wedge v^k, w^1\wedge\dots \wedge w^k\big) &\mapsto
|
|||
|
|
\det\Big(\big(g(g^\sharp(v^j),g^\sharp(w^k))\big)_{jk}\Big).
|
|||
|
|
\end{align*}
|
|||
|
|
\end{cor}
|
|||
|
|
\begin{rem}
|
|||
|
|
\label{loc-theory:rem:preserving-orthonormality}
|
|||
|
|
The two proofs of
|
|||
|
|
\Cref{loc-theory:lm:product-on-dual-space,loc-theory:lm:product-on-exterior-algebra}
|
|||
|
|
also show that the induced inner products preserve orthonormality, i.e. for an orthonormal basis
|
|||
|
|
$(v_1,\dots,v_n)$ of $V$, the induced bases $(v^1,\dots,v^n)$ of $V^*$ and $(v_J)_{J}$ of
|
|||
|
|
$\bigwedge^k V$ are also orthonormal with respect to the induced inner products. Since \Cref{loc-theory:cor:induced-product-on-exterior-algebra} just combines these inner products,
|
|||
|
|
this property also holds for the induced inner product $\tilde{g}_k$ on $\bigwedge\nolimits^kV^*$.
|
|||
|
|
\end{rem}
|
|||
|
|
|
|||
|
|
\begin{rem}
|
|||
|
|
\label{loc-theory:rem:hermitian-form-on-exterior-algebra}
|
|||
|
|
Similar to the construction of the positive definite hermitian form on $V_\mathbb{C}$, we are able
|
|||
|
|
to obtain positive definite hermitian forms $\tilde{h}_k$ on the exterior algebra spaces
|
|||
|
|
$\bigwedge\nolimits^kV_\mathbb{C}^*$, by extending the inner products $\tilde{g}_k$ sesquilinearly
|
|||
|
|
(cf. \cite[p.\,33]{Huybrechts2004}).
|
|||
|
|
\end{rem}
|
|||
|
|
|
|||
|
|
\begin{nota}
|
|||
|
|
It is common practice to only write $g$ and $h$ for all these inner products and hermitian forms,
|
|||
|
|
respectively. This is because they have similar properties and since they are all defined on different
|
|||
|
|
spaces, it should always be clear from the context which form is meant.
|
|||
|
|
\end{nota}
|
|||
|
|
|
|||
|
|
\subsection{Local operators}\;
|
|||
|
|
|
|||
|
|
The next section will focus on the definition of some essential operators. These are initially
|
|||
|
|
defined as local operators on vector spaces but are going to be used on vector bundles later.
|
|||
|
|
However, most of their properties can already be shown locally.
|
|||
|
|
|
|||
|
|
Therefore, we will assume the following setting for the remainder of this section.
|
|||
|
|
\begin{set}
|
|||
|
|
Let $(V,g, I)$ be an euclidean vector space of dimension $n$ with a compatible almost complex
|
|||
|
|
structure. Also, let $\omega$ denote the fundamental form associated with $g$ and let $h$ denote
|
|||
|
|
the induced hermitian forms on $(V,I)$ and $V_\mathbb{C}$. The existence of the almost complex
|
|||
|
|
structure $I$ ensures $n = 2m$ fore some $m \in \mathbb{N}$.
|
|||
|
|
\end{set}
|
|||
|
|
\begin{defn}[Lefschetz operator]
|
|||
|
|
With $\omega$ the associated fundamental form to $(V,g,I)$, the real \emph{Lefschetz operator} is
|
|||
|
|
defined as the linear mapping
|
|||
|
|
\begin{align*}
|
|||
|
|
L: \bigwedge\nolimits^k V^* \rightarrow \bigwedge\nolimits^{k+2} V^*,\;\enspace
|
|||
|
|
\alpha \mapsto \omega \wedge \alpha.
|
|||
|
|
\end{align*}
|
|||
|
|
The complex \emph{Lefschetz operator} on $\bigwedge\nolimits^kV^*_\mathbb{C}$ is then
|
|||
|
|
defined as the $\mathbb{C}$-linear extension $L_\mathbb{C}: \bigwedge\nolimits^k V^*_\mathbb{C}
|
|||
|
|
\rightarrow \bigwedge\nolimits^{k+2} V^*_\mathbb{C}$, i.e. $L_\mathbb{C}(\beta) = \omega \wedge
|
|||
|
|
\beta$ for all $\beta \in \bigwedge\nolimits^{k+2} V^*_\mathbb{C}$.
|
|||
|
|
\end{defn}
|
|||
|
|
\begin{rem}
|
|||
|
|
Note that $\omega \in \bigwedge\nolimits^2 V^* \cap \bigwedge\nolimits^{1,1}
|
|||
|
|
V^*\subset\bigwedge\nolimits^2V_\mathbb{C}^*$ and this is the reason for these two wedge products
|
|||
|
|
to be both meaningful. Also, we have to keep in mind that the definition of this operator depends on the
|
|||
|
|
fundamental form $\omega$, which itself depends on the choice of the inner product $g$.
|
|||
|
|
|
|||
|
|
Furthermore, due to the fundamental form $\omega$ being of type $(1,1)$, it is apparent that the
|
|||
|
|
restriction of $L_\mathbb{C}$ to forms of type $(p,q)$ behaves like
|
|||
|
|
$L: \bigwedge\nolimits^{p,q} V^* \rightarrow \bigwedge\nolimits^{p+1,q+1}V^*$
|
|||
|
|
and thus the complex Lefschetz operator preserves the revised structure of
|
|||
|
|
$\bigwedge\nolimits^k V_\mathbb{C}^* = \bigoplus_{p+q=k}\bigwedge\nolimits^{p,q}V^*$
|
|||
|
|
(cf. \cite[Proposition 1.2.8 (ii)]{Huybrechts2004}).
|
|||
|
|
\end{rem}
|
|||
|
|
|
|||
|
|
In linear algebra, we have the notion of an adjoint operator with respect to an inner product on a
|
|||
|
|
vector space. Therefore, we can use our induced inner product on the exterior algebra spaces
|
|||
|
|
$\bigwedge\nolimits^kV^*$ to define the dual of the Lefschetz operator.
|
|||
|
|
\begin{defn}
|
|||
|
|
The \emph{dual Lefschetz operator} $\Lambda$ is defined as the adjoint operator of $L$ with respect
|
|||
|
|
to the inner product $g$, i.e. the uniquely defined mapping
|
|||
|
|
\begin{align*}
|
|||
|
|
\Lambda: \bigwedge\nolimits^{k+2} V^* \rightarrow \bigwedge\nolimits^{k} V^*,
|
|||
|
|
\end{align*}
|
|||
|
|
such that for all $\alpha \in \bigwedge\nolimits^{k+2} V^*$ and $\beta \in \bigwedge\nolimits^{k}V^*$,
|
|||
|
|
it is $g(\Lambda(\alpha), \beta) = g(\alpha,L(\beta))$.
|
|||
|
|
\end{defn}
|
|||
|
|
\begin{rem}
|
|||
|
|
Note that this operator is indeed uniquely defined because of the non-degeneracy of the inner
|
|||
|
|
product $g$. This is because if we let $\alpha \in \bigwedge\nolimits^{k+2} V^*$ and assume there
|
|||
|
|
would be a second operator $\widetilde{\Lambda}$ admitting the same adjunction property, such
|
|||
|
|
that $\Lambda(\alpha) \neq \widetilde{\Lambda}(\alpha)$, it would be for all
|
|||
|
|
$\beta \in \bigwedge\nolimits^k V^*$
|
|||
|
|
\begin{align*}
|
|||
|
|
g(\Lambda(\alpha) - \widetilde{\Lambda}(\alpha), \beta) = g\left(\Lambda (\alpha), \beta\right) -
|
|||
|
|
g(\widetilde{\Lambda}(\alpha),\beta) = g(\alpha, L(\beta)) - g(\alpha, L(\beta)) = 0.
|
|||
|
|
\end{align*}
|
|||
|
|
Thus, the non-degeneracy of $g$ can be used to obtain $\Lambda (\alpha)-\widetilde{\Lambda}(\alpha) = 0$,
|
|||
|
|
which is a contradiction to our assumption.
|
|||
|
|
\end{rem}
|
|||
|
|
To properly define the next operator now, it is necessary to discuss the existence of a volume
|
|||
|
|
element in $\bigwedge^n V^*$ first.
|
|||
|
|
\begin{rem}
|
|||
|
|
\label{loc-theory:volume-form-locally}
|
|||
|
|
We already know that $V$ has a natural orientation, which we are going to call $\sigma$ for now.
|
|||
|
|
As mentioned in \cite[p.\,83 below Theorem 4-6]{Spivak1965}, there exists a unique volume form
|
|||
|
|
$\vol \in \bigwedge^nV^*$, such that $\vol(v_1, \dots, v_n)= 1$ whenever $(v_1, \dots, v_n)$ is
|
|||
|
|
an orthonormal basis of $V$ that is positively orientated with respect to $\sigma$. This volume form can
|
|||
|
|
be given as $\vol = v^1 \wedge \dots \wedge v^n$.
|
|||
|
|
|
|||
|
|
Now, we want to show that $\vol = \frac{1}{m!} \omega^m$. %with $\omega \in \bigwedge^2 V^*$
|
|||
|
|
%being the fundamental form associated to $(V,g,I)$.
|
|||
|
|
Therefore, let $(w_1,\dots,w_m)$ be a complex orthonormal basis of $(V,I)$ with respect to the
|
|||
|
|
hermitian form $h$. A simple calculation proves that the induced real basis $(w_1,I(w_1),\dots,w_m,I(w_m))$
|
|||
|
|
is orthonormal with respect to the inner product $g$ on $V$. We can calculate for all $w_j$ and $w_k$
|
|||
|
|
\begin{align*}
|
|||
|
|
\omega(w_j,I(w_k)) = g(I(w_j),I(w_k)) = g(w_j,w_k) = \delta_{jk}
|
|||
|
|
\end{align*}
|
|||
|
|
and also
|
|||
|
|
\begin{align*}
|
|||
|
|
\omega(w_j,w_k) = g(I(w_j),w_k) = 0.
|
|||
|
|
\end{align*}
|
|||
|
|
Let $I(w_j)^*$ denote the dual basis vector of $I(w_j)$. With the above calculation, we conclude
|
|||
|
|
that $\omega = \sum_{j=1}^{m} w^j \wedge I(w_j)^*$ and therefore it is
|
|||
|
|
\begin{align*}
|
|||
|
|
\omega^m = \Big(\sum_{j=1}^{m} w^j \wedge I(w_j)^*\Big)^m= m! \cdot \big(w^1 \wedge I(w_1)^*\big)
|
|||
|
|
\wedge \dots \wedge \big(w^m\wedge I(w_m)^*\big)
|
|||
|
|
= m! \cdot \vol.
|
|||
|
|
\end{align*}
|
|||
|
|
\end{rem}
|
|||
|
|
This concludes our collection of the necessary elements to define the Hodge star operator.
|
|||
|
|
\begin{defn}[Hodge star operator]
|
|||
|
|
The \emph{Hodge star operator} on $\bigwedge^k V^*$ is defined as a linear mapping
|
|||
|
|
$\hodgestar: \bigwedge^k V^* \rightarrow \bigwedge^{n-k}V^*$, such that for all
|
|||
|
|
$\alpha,\beta \in\bigwedge^kV^*$, it is
|
|||
|
|
\begin{align}
|
|||
|
|
\label{loc-theory:eq:hodge-star}
|
|||
|
|
\alpha \wedge \hodgestar\beta = g(\alpha,\beta) \cdot \vol.
|
|||
|
|
\end{align}
|
|||
|
|
The $(n-k)$ form $\hodgestar \beta$ is called the \emph{Hodge dual} of $\beta$.
|
|||
|
|
\end{defn}
|
|||
|
|
Note that \Cref{loc-theory:eq:hodge-star} uniquely defines the Hodge dual because for $r,s \in
|
|||
|
|
\mathbb{N}$ with $r + s = n$, the exterior product defines a non-degenerate pairing
|
|||
|
|
\begin{align*}
|
|||
|
|
\bigwedge\nolimits^r V^* \times \bigwedge\nolimits^s V^* \rightarrow \bigwedge\nolimits^{n} V^*.
|
|||
|
|
\end{align*}
|
|||
|
|
|
|||
|
|
To show that such an operator exists, we choose an orthonormal basis $(v_1,\dots,v_n)$ of
|
|||
|
|
$V$ that is positively oriented with respect to the natural orientation of $V$. With
|
|||
|
|
\Cref{loc-theory:rem:preserving-orthonormality}, we already know that the induced basis $(v^J)_{J}$ of
|
|||
|
|
$\bigwedge^kV^*$ is orthonormal as well. For an arbitrary permutation $\tau \in S_n$, we set the Hodge star
|
|||
|
|
operator to map as follows:
|
|||
|
|
\begin{align}
|
|||
|
|
\label{loc-theory:eq:explicit-hodge-star}
|
|||
|
|
\hodgestar\big(v^{\tau(1)}\wedge\dots\wedge v^{\tau(k)}\big) = \sign(\tau) \cdot \;v^{\tau(k+1)}
|
|||
|
|
\wedge\dots\wedge v^{\tau(n)}.
|
|||
|
|
\end{align}
|
|||
|
|
With this definition, it is
|
|||
|
|
\begin{align*}
|
|||
|
|
v^{\tau(1)}\wedge\dots\wedge v^{\tau(k)} \wedge \hodgestar\big(v^{\tau(1)}\wedge\dots\wedge v^{\tau(k)}\big) &= \sign(\tau) \cdot v^{\tau(1)} \wedge\dots\wedge v^{\tau(n)}\\
|
|||
|
|
&= \sign(\tau)^2\cdot v^1\wedge \dots \wedge v^n \\&= 1 \cdot \vol.
|
|||
|
|
\end{align*}
|
|||
|
|
At the same time, it is for all $v^{j_1}\wedge\dots\wedge v^{j_k} \neq \pm v^{\tau(1)}\wedge\dots\wedge v^{\tau(k)}$
|
|||
|
|
\begin{align*}
|
|||
|
|
v^{j_1}\wedge\dots\wedge v^{j_k} \wedge \hodgestar\big(v^{\tau(1)}\wedge\dots\wedge
|
|||
|
|
v^{\tau(k)}\big) = 0.
|
|||
|
|
\end{align*} This is because if we assume without loss of generality that $j_1\neq\dots\neq j_k$,
|
|||
|
|
then there exists at least one $s\in\mathbb{N}$ with $(k+1) \leq s \leq n$, such that $\tau(s) \in \{j_1,\dots,j_k\}$.
|
|||
|
|
This shows that the above mapping indeed explicitly defines the Hodge star. See also \cite[p.\,56f]{Schnell2012},
|
|||
|
|
where we have later discovered a very similar calculation.
|
|||
|
|
\begin{prop}[Properties of the Hodge star {\cite[Proposition 1.2.20]{Huybrechts2004}}]\;\\
|
|||
|
|
\label{loc-theory:lm:property-hodge-star}
|
|||
|
|
Among others, the Hodge star operator has the following properties:
|
|||
|
|
\begin{enumerate}
|
|||
|
|
\item The Hodge star operator on $\bigwedge^kV^*$ is an isometric isomorphism, i.e. it is
|
|||
|
|
bijective and for all $\alpha, \beta \in \bigwedge^kV^*$, it is $g(\alpha,\beta) = g(\hodgestar\alpha,\hodgestar\beta)$.
|
|||
|
|
\item For all $\alpha \in \bigwedge^kV^*$, it is $\hodgestar^2\alpha = (-1)^k \alpha$. In particular, it is
|
|||
|
|
$\hodgestar^{-1} = (-1)^k \hodgestar$.
|
|||
|
|
\label{loc-theory:lm:property-hodge-star-one}
|
|||
|
|
\end{enumerate}
|
|||
|
|
\end{prop}
|
|||
|
|
\begin{proof}
|
|||
|
|
With the explicit definition in \Cref{loc-theory:eq:explicit-hodge-star}, it is quite easy to determine
|
|||
|
|
that the Hodge star operator maps any orthonormal basis to an orthonormal basis, and this proves
|
|||
|
|
property (1).
|
|||
|
|
For the proof of property (2), we use a local version of the calculation in \cite[Lemma 5.5]{Voisin2002}.
|
|||
|
|
We can apply property (1) and calculate for all $\alpha,\beta \in\bigwedge^kV^*$
|
|||
|
|
\begin{align*}
|
|||
|
|
\alpha\wedge\hodgestar\beta = g(\alpha, \beta) \cdot \vol = g(\hodgestar\alpha,\hodgestar\beta)
|
|||
|
|
\cdot \vol = g(\hodgestar\beta,\hodgestar\alpha) \cdot \vol = \hodgestar\beta \wedge \hodgestar^2
|
|||
|
|
\alpha.
|
|||
|
|
\end{align*}
|
|||
|
|
With $n-(n-k) = k$, we know $\hodgestar^2 \alpha$ is a $k$-form again. As $\hodgestar\beta$ is a
|
|||
|
|
$(n-k)$ form, we obtain
|
|||
|
|
\begin{align*}
|
|||
|
|
\hodgestar\beta \wedge \hodgestar^2\alpha &= (-1)^{k(n-k)} \hodgestar^2 \alpha \wedge \hodgestar
|
|||
|
|
\beta \\&= (-1)^{k(2m-k)} \hodgestar^2\alpha\wedge\hodgestar\beta\\ &= (-1)^k \hodgestar^2 \alpha
|
|||
|
|
\wedge \hodgestar\beta.
|
|||
|
|
\end{align*}
|
|||
|
|
Given that this holds for all $\beta \in \bigwedge^kV^*$, the second property is already proven.
|
|||
|
|
\end{proof}
|
|||
|
|
Additionally, there is an interesting relation between the Hodge star operator and the Lefschetz and
|
|||
|
|
dual Lefschetz operators, which is going to be useful later.
|
|||
|
|
\begin{lm}[{\cite[Lemma 1.2.23]{Huybrechts2004}}]
|
|||
|
|
\label{loc-theory:lm:formula-for-the-dual-lefschetz-operator}
|
|||
|
|
For $\alpha \in \bigwedge^{k+2} V^*$ the image under the dual Lefschetz operator $\Lambda$ can be
|
|||
|
|
explicitly calculated as
|
|||
|
|
\begin{align*}
|
|||
|
|
\Lambda (\alpha) = \big(\hodgestar^{-1} \circ \,L \circ \hodgestar\big)(\alpha) = \big( (-1)^k
|
|||
|
|
\hodgestar \circ \,L \circ \hodgestar \big)(\alpha).
|
|||
|
|
\end{align*}
|
|||
|
|
\end{lm}
|
|||
|
|
\begin{proof} We expand the proof of the given lemma in \cite{Huybrechts2004}.
|
|||
|
|
Let $\beta \in \bigwedge^kV^*$. Using the definition of the Hodge star and the definition of the
|
|||
|
|
Lefschetz operator, we can calculate
|
|||
|
|
\begin{align*}
|
|||
|
|
g(\alpha, L\beta)\cdot \vol = g(L\beta,\alpha) \cdot \vol = L\beta \wedge \hodgestar \alpha =
|
|||
|
|
\omega \wedge \beta \wedge \hodgestar \alpha.
|
|||
|
|
\end{align*}
|
|||
|
|
As $\omega$ is a $2$-form and the wedge product is associative, we have
|
|||
|
|
\begin{align*}
|
|||
|
|
g(\alpha, L\beta)\cdot \vol = (-1)^{2k} \beta \wedge \omega \wedge \hodgestar\alpha = \beta
|
|||
|
|
\wedge (\omega \wedge \hodgestar\alpha).
|
|||
|
|
\end{align*}
|
|||
|
|
Applying the definition of the Lefschetz operator again and using the definition of the Hodge star
|
|||
|
|
operator yields
|
|||
|
|
\begin{align*}
|
|||
|
|
g(\alpha, L\beta)\cdot \vol &=\beta \wedge L (\hodgestar \alpha) \\
|
|||
|
|
&= \beta \wedge (\hodgestar \hodgestar^{-1}) L (\hodgestar \alpha) \\
|
|||
|
|
&= \beta \wedge \hodgestar (\hodgestar^{-1} (L (\hodgestar \alpha))) \\
|
|||
|
|
&= g\big(\beta, (\hodgestar^{-1}( L (\hodgestar \alpha)))\big) \cdot \vol \\
|
|||
|
|
&=g(\hodgestar^{-1} (L (\hodgestar \alpha)), \beta)\cdot\vol.
|
|||
|
|
\end{align*}
|
|||
|
|
Thus, we have shown the equality
|
|||
|
|
$g(\Lambda \alpha, \beta) = g(\alpha,L\beta) = g(\hodgestar^{-1}(L(\hodgestar \alpha)), \beta)$.
|
|||
|
|
Since this holds for all $k$-forms $\beta$, the non-degeneracy of $g$ proves the first equality
|
|||
|
|
of the statement. The second equality then follows directly with
|
|||
|
|
\creflmpart{loc-theory:lm:property-hodge-star}{loc-theory:lm:property-hodge-star-one}. % see custom command in main_thesis
|
|||
|
|
\end{proof}
|
|||
|
|
Similar to the $\mathbb{C}$-linear extension of the Lefschetz operator, we will also need the
|
|||
|
|
$\mathbb{C}$-linear extension of the Hodge star operator.
|
|||
|
|
\begin{defn}
|
|||
|
|
The $\mathbb{C}$-linear extension of the Hodge star
|
|||
|
|
$\hodgestar_\mathbb{C} :\bigwedge^kV_\mathbb{C}^* \rightarrow \bigwedge^{n-k} V_\mathbb{C}^*$
|
|||
|
|
is defined such that for all $\alpha,\beta \in \bigwedge^kV^*_\mathbb{C}$, we have
|
|||
|
|
\begin{align*}
|
|||
|
|
\alpha \wedge \hodgestar_\mathbb{C}\overline{\beta} = h(\alpha, \beta) \cdot \vol.
|
|||
|
|
\end{align*}
|
|||
|
|
Note that this expression is meaningful because
|
|||
|
|
$\vol = \frac{1}{m!} \omega^m \in \bigwedge^nV^*\cap \bigwedge^{m,m}V^*$.
|
|||
|
|
Since the hermitian form $h$ on $\bigwedge^kV_\mathbb{C}^*$ was previously defined as the
|
|||
|
|
sesquilinear extension of the inner product $g$ on $\bigwedge^kV^*$, it is immediate that this is indeed
|
|||
|
|
the $\mathbb{C}$-linear extension of the real Hodge star operator.
|
|||
|
|
\end{defn}
|
|||
|
|
There is also the $\mathbb{C}$-linear extension
|
|||
|
|
$\Lambda_\mathbb{C}: \bigwedge^{k+2}V_\mathbb{C}^*\rightarrow \bigwedge^{k} V_\mathbb{C}^*$
|
|||
|
|
of the dual Lefschetz operator. Using the explicit formula in
|
|||
|
|
\Cref{loc-theory:lm:formula-for-the-dual-lefschetz-operator}, this extension is given as
|
|||
|
|
\begin{align*}
|
|||
|
|
\Lambda_\mathbb{C} = (-1)^k \hodgestar_\mathbb{C} \circ L_\mathbb{C} \circ \hodgestar_\mathbb{C}.
|
|||
|
|
\end{align*}
|
|||
|
|
We can use the same calculation as in the proof of \Cref{loc-theory:lm:formula-for-the-dual-lefschetz-operator}
|
|||
|
|
to show that this is indeed the adjoint operator to $L$ with respect to the hermitian form $h$.
|
|||
|
|
\begin{nota}
|
|||
|
|
It is common practice to abuse the notation, denoting the complex extensions
|
|||
|
|
$L_\mathbb{C},\hodgestar_\mathbb{C}$ and $\Lambda_\mathbb{C}$ as $L, \hodgestar$
|
|||
|
|
and $\Lambda$, respectively.
|
|||
|
|
\end{nota}
|