Almost complex structure at fiber level. A module structure on $\mathbb{R}^{m}$ over $\mathbb{R}[t]/(t^{2} + 1) = \mathbb{C}$ is precisely a matrix $J \in \mathrm{Mat}_{m}(\mathbb{R})$ such that $J^{2} + \mathrm{id} = 0.$ We call such $J$ an almost complex structure on $\mathbb{R}^{m}.$
Remark. Existence of such $J$ implies that $m$ is even by considering the $\mathbb{R}[t]$-module structure given on $\mathbb{R}^{m}$ due to $J.$ Hence, we may write $m = 2n.$
We may consider $$J_{\mathbb{C}} := J \otimes_{\mathbb{R}} \mathrm{id} : \mathbb{R}^{2n} \otimes_{\mathbb{R}} \mathbb{C} \rightarrow \mathbb{R}^{2n} \otimes_{\mathbb{R}} \mathbb{C},$$ given by $v \otimes c \mapsto J(v) \otimes c.$ Identifying $\mathbb{R}^{2n} \otimes_{\mathbb{R}} \mathbb{C}$ with $\mathbb{C}^{2n},$ we can consider $J_{\mathbb{C}} \in \mathrm{Mat}_{2n}(\mathbb{C}),$ and as a matrix it can be thought of just $J$ but as the image of the embedding $$\mathrm{Mat}_{2n}(\mathbb{R}) \hookrightarrow \mathrm{Mat}_{2n}(\mathbb{C})$$ given by $$\mathbb{R} \hookrightarrow \mathbb{C}.$$ Hence, we will just write $J$ without $\mathbb{C}$ in the notation. We note that $J$ gives $\mathbb{C}^{2n}$ a module structure over $$\mathbb{C}[t]/(t^{2} + 1) \simeq \mathbb{C}[t]/(t - i) \times \mathbb{C}[t]/(t + i),$$ and this implies that we have a decomposition $$\mathbb{C}^{2n} = V^{1,0} \oplus V^{0,1},$$ where on $V^{1,0},$ the linear map $J$ acts as the multiplication by $i,$ while on $V^{0,1},$ the linear map $J$ acts as the multiplication by $-i.$ We write $Q : \mathbb{C}^{2n} \rightarrow \mathbb{C}^{2n}$ to denote the map given by complex conjugating each entry. (Not that important: $\mathbb{R}^{2n}$ is the set of points in $\mathbb{C}^{2n}$ fixed by $Q.$)
Lemma. We continue to assume that there is $J \in \mathrm{Mat}_{2n}(\mathbb{R})$ with $J^{2} + \mathrm{id} = 0.$ The map $\mathbb{R}^{2n} \hookrightarrow \mathbb{C}^{2n} \twoheadrightarrow V^{1,0}$ is an isomorphism of $\mathbb{C}$-vector spaces, while the map $\mathbb{R}^{2n} \hookrightarrow \mathbb{C}^{2n} \twoheadrightarrow V_{0,1}$ is a conjugate isomorphism.
Proof. Since $J^{2} + \mathrm{id} = 0,$ the $\mathbb{C}[t]$-module structure on $\mathbb{C}^{2n}$ given by $J$ is necessarily of the form $$\mathbb{C}^{2n} \simeq (\mathbb{C}[t]/(t^{2} + 1))^{\oplus n} \simeq (\mathbb{C}[t]/(t - i))^{\oplus n} \oplus (\mathbb{C}[t]/(t + i))^{\oplus n}.$$ Hence, we have $n = \dim_{\mathbb{C}}(V^{1,0}) = \dim_{\mathbb{C}}(V^{0,1}).$ This implies that $$2n = \dim_{\mathbb{R}}(V^{1,0}) = \dim_{\mathbb{R}}(V^{0,1}).$$ Fix any $v \in \mathbb{R}^{2n}.$ Then we have $$v = v^{1,0} + v^{0,1} \in \mathbb{C}^{n}$$ with $v^{1,0} \in V^{1,0}$ and $v^{0,1} \in V^{0,1}.$ Note that $J(v) = i v^{1,0} - i v^{0,1}$ and thus $iJ(v) = -v^{1,0} + v^{0,1}.$ This implies that $$v - iJ(v) = 2v^{1,0}$$ and $$v + iJ(v) = 2v^{0,1}.$$ Since $v \in \mathbb{R}^{2n},$ we have $J(v) \in \mathbb{R}^{2n}.$ Thus, if $v^{1,0},$ then $v = 0.$ Likewise, if $v^{0,1} = 0,$ then $v = 0.$ This implies that the maps in question are both injective. Since they are $\mathbb{R}$-linear maps, by dimension counting they are $\mathbb{R}$-isomoprhisms. The first map is given by $f^{1,0} : v \mapsto v^{1,0},$ so $$f^{1,0}(iv) = f^{1,0}(J(v)) = f^{1,0}(iv^{1,0} - iv^{0,1}) = iv^{1,0} = if^{1,0}(v).$$ Hence $f^{1,0}$ is $\mathbb{C}$-linear. Likewise, the map $f^{0,1} : v \mapsto v^{0,1}$ is conjugate linear. This finishes the proof. $\Box$
Remark. In the proof above, given any $v \in \mathbb{R}^{2n},$ we realized the following explicit formulas: $$v^{1,0} = \frac{v - iJ(v)}{2}$$ and $$v^{0,1} = \frac{v + iJ(v)}{2}.$$ Apparently this does not work in characteristic $2,$ but isn't it okay because we did not divide by $2$ in the proof? Characteristic $2$ does not work even before that, because we have $t^{2} + 1 = (t + 1)^{2}$ over any characteristic $2$ field. This explicit description lets us see the following.
Corollary. Keeping the above notations, we have $\overline{V^{1,0}} = V^{0,1},$ where $\overline{W} := Q(W).$
Dual. Given any $A \in \mathrm{Mat}_{m}(\mathbb{R}),$ we can let $A$ act on $(\mathbb{R}^{m})^{\vee} = \mathrm{Hom}_{\mathbb{R}}(\mathbb{R}^{m}, \mathbb{R})$ by defining $(A \varphi)(v) := \varphi(Av).$ This imbues $(\mathbb{R}^{m})^{\vee}$ an $\mathbb{R}[t]$-module structure.
Any $J \in \mathrm{Mat}_{2n}(\mathbb{R})$ with $J^{2} + \mathrm{id} = 0,$ gives $(\mathbb{R}^{2n})^{\vee}$ a module structure over $\mathbb{R}[t]/(t^{2} + 1) = \mathbb{C}.$ Thus, we get a decomposition $$(\mathbb{R}^{2n})^{\vee} \otimes_{\mathbb{R}} \mathbb{C} = U^{1,0} \oplus U^{0,1}$$ like before.
Any $J \in \mathrm{Mat}_{2n}(\mathbb{R})$ with $J^{2} + \mathrm{id} = 0,$ gives $(\mathbb{R}^{2n})^{\vee}$ a module structure over $\mathbb{R}[t]/(t^{2} + 1) = \mathbb{C}.$ Thus, we get a decomposition $$(\mathbb{R}^{2n})^{\vee} \otimes_{\mathbb{R}} \mathbb{C} = U^{1,0} \oplus U^{0,1}$$ like before.
Exterior power. We also have $$\begin{align*}\bigwedge^{r} ((\mathbb{R}^{2n})^{\vee} \otimes_{\mathbb{R}} \mathbb{C}) &= \bigwedge^{r} (U^{1,0} \oplus U^{0,1}) \\ &= \bigoplus_{\substack{p, q \geq 1 \\ p+q = r } } \left( \bigwedge^{p} U^{1,0} \otimes_{\mathbb{R}} \bigwedge^{q} U^{0,1} \right),\end{align*}$$ and by looking element-wise, the complex conjugation map $Q$ induces a conjugation map on this $r$-th wedge product that maps $\bigwedge^{p} U^{1,0} \otimes_{\mathbb{R}} \bigwedge^{q} U^{0,1}$ to $\bigwedge^{q} U^{1,0} \otimes_{\mathbb{R}} \bigwedge^{p} U^{0,1}.$
Description of basis at fiber level. Let $V$ be a finite dimensional real vector space with an almost complex structure $J,$ so that in particular $\dim_{\mathbb{R}}(V) = 2n.$ This gives $$V_{\mathbb{C}} := V \otimes_{\mathbb{R}} \mathbb{C} = V^{1,0} \oplus V^{0,1},$$ which is $2n$-dimensional complex vector space. Let $x_{1}, \dots, x_{n}, y_{1}, \dots, y_{n}$ be a basis for $V.$ We already know from previous computations that writing $$z_{j} = \frac{x_{j} - iy_{j}}{2}$$ and $$\bar{z_{j}} = \frac{x_{j} + iy_{j}}{2},$$ we have $$V^{1,0} = \mathbb{C}z_{1} \oplus \cdots \oplus \mathbb{C}z_{n}$$ and $$V^{0,1} = \mathbb{C}\bar{z_{1}} \oplus \cdots \oplus \mathbb{C}\bar{z_{n}}.$$ Let $x_{1}^{\vee}, \dots, x_{n}^{\vee}, y_{1}^{\vee}, \dots, y_{n}^{\vee}$ the basis of $$V^{\vee} = \mathrm{Hom}_{\mathbb{R}}(V, \mathbb{R})$$ dual to $x_{1}, \dots, x_{n}, y_{1}, \dots, y_{n}.$ We have $$V^{\vee} \otimes_{\mathbb{R}} \mathbb{C} = U^{1,0} \oplus U^{0,1}.$$ We will make use of the following isomorphism: $$\mathrm{Hom}_{\mathbb{R}}(V, \mathbb{R}) \otimes_{\mathbb{R}} \mathbb{C} \simeq \mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$$ determined by $v^{\vee} \otimes 1 \mapsto (v \otimes 1)^{\vee}$ for any $v \in V,$ where $0^{\vee} = 0$ and $v^{\vee}$ is the map sending $v \mapsto 1$ and everything else to $0.$ This isomorphism is useful in getting our hands on $U^{1,0}$ and $U^{0,1}.$ This is because the right-hand side gives us $$\mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}) = \mathrm{Hom}_{\mathbb{C}}(V^{1,0} \oplus V^{0,1}, \mathbb{C}) \simeq \mathrm{Hom}_{\mathbb{C}}(V^{1,0}, \mathbb{C}) \times \mathrm{Hom}_{\mathbb{C}}(V^{0,1}, \mathbb{C}).$$ We have:
Lemma. Under the isomorphism $$\mathrm{Hom}_{\mathbb{R}}(V, \mathbb{R}) \otimes_{\mathbb{R}} \mathbb{C} \simeq \mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}),$$ we have $$U^{1,0} \simeq \mathrm{Hom}_{\mathbb{C}}(V^{1,0}, \mathbb{C})$$ and $$U^{0,1} \simeq \mathrm{Hom}_{\mathbb{C}}(V^{0,1}, \mathbb{C}),$$ and the first takes the action of $J$ as the multiplication by $i$ and the second the multiplication by $-i.$
Proof. Denote by $\phi$ the given isomorphism, so $$\phi(v^{\vee} \otimes 1) = (v \otimes 1)^{\vee}$$ for any basis element $v$ of $V.$ We have $$\begin{align*}\phi(J(v^{\vee} \otimes 1)) &= \phi((Jv)^{\vee} \otimes 1) \\ &= (Jv \otimes 1)^{\vee} \\ &= (J(v \otimes 1))^{\vee} \\ &= J(v \otimes 1)^{\vee}.\end{align*}$$ Hence, to show the claim it is enough to show that the $i$-eigenspace of the action of $J$ on $\mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$ is $\mathrm{Hom}_{\mathbb{C}}(V^{1,0}, \mathbb{C}),$ while the $(-i)$-eigenspace is $\mathrm{Hom}_{\mathbb{C}}(V^{0,1}, \mathbb{C}).$ Given $$(f^{1,0}, 0) \in \mathrm{Hom}_{\mathbb{C}}(V^{1,0}, \mathbb{C}) \hookrightarrow \mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}),$$ we have $$\begin{align*}(J \cdot (f^{1,0}, 0)) (v \otimes 1) &= (f^{1,0}, 0) (Jv \otimes 1) \\ &= (f^{1,0}, 0) (iv^{1,0} - iv^{1,0}) \\ &= if^{1,0}(v^{1,0}) \\ &= (i \cdot (f^{1,0}, 0))(v \otimes 1), \end{align*}$$ so we have $J (f^{1,0}, 0) = i (f^{1,0}, 0).$ Likewise, we can deduce $J (0, f^{0,1}) = -i (0, f^{0,1})$ for any $$(0, f^{0,1}) \in \mathrm{Hom}_{\mathbb{C}}(V^{0,1}, \mathbb{C}) \hookrightarrow \mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}).$$ This finishes the proof. $\Box$
Note that $x_{1}^{\vee} + iy_{1}^{\vee}, \dots, x_{n} + iy_{n}^{\vee}, x_{1}^{\vee} - iy_{1}^{\vee}, \dots, x_{n} - iy_{n}^{\vee} \in \mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$ form a dual basis for $z_{1}, \dots, z_{n}, \bar{z_{1}}, \dots, \bar{z_{n}} \in V \otimes_{\mathbb{R}} \mathbb{C}.$
Lemma. We have $$x_{1}^{\vee} + iy_{1}^{\vee}, \dots, x_{n} + iy_{n}^{\vee} \in \mathrm{Hom}_{\mathbb{C}}(V^{1,0} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$$ and $$x_{1}^{\vee} - iy_{1}^{\vee}, \dots, x_{n} - iy_{n}^{\vee} \in \mathrm{Hom}_{\mathbb{C}}(V^{0,1} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}).$$ In particular, these elements form $\mathbb{C}$-bases for $\mathrm{Hom}_{\mathbb{C}}(V^{1,0} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$ and $\mathrm{Hom}_{\mathbb{C}}(V^{0,1} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}).$
Proof. Fix a general element $$w = a_{1}z_{1} + \cdots + a_{n}z_{n} + b_{1}\bar{z_{1}} + \cdots + b_{n}\bar{z_{n}} \in V \otimes_{\mathbb{R}} \mathbb{C}.$$ We check $$\begin{align*}(J (x_{j}^{\vee} + iy_{j}^{\vee}))(w) &= (x_{j}^{\vee} + iy_{j}^{\vee})(ia_{j}z_{j}) \\ &= (i(x_{j}^{\vee} + iy_{j}^{\vee}))(a_{j}z_{j}) \\ &= (i(x_{j}^{\vee} + iy_{j}^{\vee}))(w)\end{align*}$$ and $$\begin{align*}(J (x_{j}^{\vee} - iy_{j}^{\vee}))(w) &= (x_{j}^{\vee} + iy_{j}^{\vee})(-ib_{j}\bar{z}_{j}) \\ &= (-i(x_{j}^{\vee} - iy_{j}^{\vee}))(b_{j}\bar{z}_{j}) \\ &= (i(x_{j}^{\vee} - iy_{j}^{\vee}))(w).\end{align*}$$ This implies that $J (x_{j}^{\vee} + iy_{j}^{\vee}) = i (x_{j}^{\vee} + iy_{j}^{\vee})$ so that $$x_{j}^{\vee} + iy_{j}^{\vee} \in \mathrm{Hom}_{\mathbb{C}}(V^{1,0} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$$ due to the previous lemma. Similarly, we get $$x_{j}^{\vee} - iy_{j}^{\vee} \in \mathrm{Hom}_{\mathbb{C}}(V^{0,1} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}),$$ as desired. $\Box$
What's the upshot? Again, we have $$V^{\vee} \otimes_{\mathbb{R}} \mathbb{C} \simeq \mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$$ determined by $v^{\vee} \otimes 1 \mapsto (v \otimes 1)^{\vee},$ which gives $$U^{1,0} \simeq \mathrm{Hom}_{\mathbb{C}}(V^{1,0} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$$ and $$U^{0,1} \simeq \mathrm{Hom}_{\mathbb{C}}(V^{0,1} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$$ the $\mathbb{C}$-bases we found for the right-hand sides produce those of the left-hand sides. Moreover, for any pure tensors $f \otimes a \in V^{\vee} \otimes_{\mathbb{R}} \mathbb{C}$ and $v \otimes b \in V \otimes_{\mathbb{R}} \mathbb{C},$ we can make sense of the evaluation $$(f \otimes a)(v \otimes b) := f(v) \otimes ab \in \mathbb{C},$$ which gives a definition of evaluation of any element of $V \otimes_{\mathbb{R}} \mathbb{C}$ by any element of $V^{\vee} \otimes_{\mathbb{R}} \mathbb{C}.$
Globalization. Let $M$ be a smooth real manifold and $E$ be a smooth real vector bundle on $M.$ An almost complex structure on $E$ is a map $J : E \rightarrow E$ of vector bundles such that $J^{2} = -\mathrm{id}.$ (This makes sense fiber-wise, but it seem okay globally as well.) Note that $J$ will force $E$ have an even rank, say $2r.$ The above discussion globalizes, so we have $$E_{\mathbb{C}} := E \otimes_{\mathbb{R}} \mathbb{C} = E^{1,0} \oplus E^{0,1},$$ coming from fiber-wise description. Everything we discussed in at fiber level works.
What do we mean by tensor product? Write $n = \mathrm{rk}(E)$ for now. We can describe $E$ by various smooth transition functions of the bundle $$(U \cap V) \times \mathbb{R}^{n} \simeq (U \cap V) \times \mathbb{R}^{n},$$ where $U, V$ are from a trivializing chart, which can be presented by smooth maps $U \cap V \rightarrow \mathrm{GL}_{n}(\mathbb{R}).$ We can then consider the smooth maps given by $U \cap V \rightarrow \mathrm{GL}_{n}(\mathbb{R}) \hookrightarrow \mathrm{GL}_{n}(\mathbb{C}),$ which will give us smooth transition functions $$(U \cap V) \times \mathbb{C}^{n} \simeq (U \cap V) \times \mathbb{C}^{n},$$ and these glue back to give $E_{\mathbb{C}}.$
Almost complex structure on tangent bundles. Let $M$ be a smooth real manifold. An almost complex structure on $M$ is an almost complex structure on its tangent bundle $TM$ (i.e., it is a map $J : TM \rightarrow TM$ of bundles with $J^{2} = -\mathrm{id}.$) We write $(TM)_{\mathbb{C}} = T^{1,0}M \oplus T^{0,1}M$ to mean that $T^{1,0}M$ is where $J$ acts as the multiplication by $i,$ while $T^{0,1}M$ by $-i.$ (We will use this notation for other vector bundles and vector spaces from now on.) Note that if $M$ is a complex manifold, then $TM,$ as a real bundle, is already equipped with the complex structure given by multiplication of $i.$
Proposition. Let $M$ be a complex manifold of dimension $n.$ Then there is an almost complex structure on $M$ (or $TM$).
Proof. It is enough to treat the case of (nonempty) open subsets of $\mathbb{C}^{n}$ and then show that biholomorphic maps preserve this structure. Fix any nonempty open $U \subset \mathbb{C}^{n},$ with complex coordinates $z_{j} = x_{j} + i y_{j}$ with $1 \leq j \leq n.$ Then $TU$ has frame (i.e., global sections $U \rightarrow TU$ giving a real basis at each fiber) given by $$\frac{\partial}{\partial x_{1}}, \dots, \frac{\partial}{\partial x_{n}}, \frac{\partial}{\partial y_{1}}, \dots, \frac{\partial}{\partial y_{n}},$$ which gives a trivialization. Define $J : TU \rightarrow TU$ by $$\frac{\partial}{\partial x_{j}} \mapsto \frac{\partial}{\partial y_{j}}$$ and $$\frac{\partial}{\partial y_{j}} \mapsto -\frac{\partial}{\partial x_{j}}.$$ Since $J^{2} = -\mathrm{id},$ this $J$ is an almost complex structure on $TU.$ Hence, we get a decomposition $$TU = T^{1,0}U \oplus T^{0,1}U,$$ where the first component is trivialized by the frame $$\frac{\partial}{\partial z_{1}}, \dots, \frac{\partial}{\partial z_{n}},$$ while the second component is trivialized by the frame $$\frac{\partial}{\partial \bar{z_{1}}}, \dots, \frac{\partial}{\partial \bar{z_{n}}}.$$ Let $V \subset \mathbb{C}^{n}$ be another nonempty open subset. In an earlier lecture, we showed that given any holomorphic map $\phi : U \rightarrow V,$ the map $(TU)_{\mathbb{C}} \rightarrow (\phi^{*}TV)_{\mathbb{C}}$ (which fiber-wise looks like $T_{x}U \otimes_{\mathbb{R}} {\mathbb{C}} \rightarrow T_{\phi(x)}V \otimes_{\mathbb{R}} {\mathbb{C}}$) restricts to $$T^{1,0}U \rightarrow \phi^{*}T^{1,0}V$$ and $$T^{0,1}U \rightarrow \phi^{*}T^{0,1}V.$$ Recall that this actually uses the fact that $\phi$ is holomorphic. In particular, if $\phi$ is biholomorphic, then all the maps above will be isomorphisms of vector bundles preserving $\mathbb{C}$-linearity fiber-wise. $\Box$
Remark. The almost complex structure given on a complex manifold in the proof above is said to be canonical.
Corollary. Let $M$ be a complex manifold of dimension $n.$ Give $M$ the canonical almost complex structure so that we can have the decomposition $$(TM)_{\mathbb{C}} = T^{1,0}M \oplus T^{0,1}M.$$ Then $T^{1,0}M$ is a holomorphic vector bundle.
Proof. It is enough to show that given two nonempty subsets $U, V \subset \mathbb{C}^{n}$ and a holomorphic map $\phi : U \rightarrow V,$ the induced map $T^{1,0}U \rightarrow \phi^{*}T^{1,0}V$ is holomorphic. The induce map $T^{1,0}U \rightarrow T^{1,0}V$ is given by the matrix $$\left[ \frac{\partial \phi_{j}}{\partial z_{k}} \right]_{1 \leq j,k \leq n},$$ and thus each entry is holomorphic. $\Box$
Description of basis at fiber level. Let $V$ be a finite dimensional real vector space with an almost complex structure $J,$ so that in particular $\dim_{\mathbb{R}}(V) = 2n.$ This gives $$V_{\mathbb{C}} := V \otimes_{\mathbb{R}} \mathbb{C} = V^{1,0} \oplus V^{0,1},$$ which is $2n$-dimensional complex vector space. Let $x_{1}, \dots, x_{n}, y_{1}, \dots, y_{n}$ be a basis for $V.$ We already know from previous computations that writing $$z_{j} = \frac{x_{j} - iy_{j}}{2}$$ and $$\bar{z_{j}} = \frac{x_{j} + iy_{j}}{2},$$ we have $$V^{1,0} = \mathbb{C}z_{1} \oplus \cdots \oplus \mathbb{C}z_{n}$$ and $$V^{0,1} = \mathbb{C}\bar{z_{1}} \oplus \cdots \oplus \mathbb{C}\bar{z_{n}}.$$ Let $x_{1}^{\vee}, \dots, x_{n}^{\vee}, y_{1}^{\vee}, \dots, y_{n}^{\vee}$ the basis of $$V^{\vee} = \mathrm{Hom}_{\mathbb{R}}(V, \mathbb{R})$$ dual to $x_{1}, \dots, x_{n}, y_{1}, \dots, y_{n}.$ We have $$V^{\vee} \otimes_{\mathbb{R}} \mathbb{C} = U^{1,0} \oplus U^{0,1}.$$ We will make use of the following isomorphism: $$\mathrm{Hom}_{\mathbb{R}}(V, \mathbb{R}) \otimes_{\mathbb{R}} \mathbb{C} \simeq \mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$$ determined by $v^{\vee} \otimes 1 \mapsto (v \otimes 1)^{\vee}$ for any $v \in V,$ where $0^{\vee} = 0$ and $v^{\vee}$ is the map sending $v \mapsto 1$ and everything else to $0.$ This isomorphism is useful in getting our hands on $U^{1,0}$ and $U^{0,1}.$ This is because the right-hand side gives us $$\mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}) = \mathrm{Hom}_{\mathbb{C}}(V^{1,0} \oplus V^{0,1}, \mathbb{C}) \simeq \mathrm{Hom}_{\mathbb{C}}(V^{1,0}, \mathbb{C}) \times \mathrm{Hom}_{\mathbb{C}}(V^{0,1}, \mathbb{C}).$$ We have:
Lemma. Under the isomorphism $$\mathrm{Hom}_{\mathbb{R}}(V, \mathbb{R}) \otimes_{\mathbb{R}} \mathbb{C} \simeq \mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}),$$ we have $$U^{1,0} \simeq \mathrm{Hom}_{\mathbb{C}}(V^{1,0}, \mathbb{C})$$ and $$U^{0,1} \simeq \mathrm{Hom}_{\mathbb{C}}(V^{0,1}, \mathbb{C}),$$ and the first takes the action of $J$ as the multiplication by $i$ and the second the multiplication by $-i.$
Proof. Denote by $\phi$ the given isomorphism, so $$\phi(v^{\vee} \otimes 1) = (v \otimes 1)^{\vee}$$ for any basis element $v$ of $V.$ We have $$\begin{align*}\phi(J(v^{\vee} \otimes 1)) &= \phi((Jv)^{\vee} \otimes 1) \\ &= (Jv \otimes 1)^{\vee} \\ &= (J(v \otimes 1))^{\vee} \\ &= J(v \otimes 1)^{\vee}.\end{align*}$$ Hence, to show the claim it is enough to show that the $i$-eigenspace of the action of $J$ on $\mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$ is $\mathrm{Hom}_{\mathbb{C}}(V^{1,0}, \mathbb{C}),$ while the $(-i)$-eigenspace is $\mathrm{Hom}_{\mathbb{C}}(V^{0,1}, \mathbb{C}).$ Given $$(f^{1,0}, 0) \in \mathrm{Hom}_{\mathbb{C}}(V^{1,0}, \mathbb{C}) \hookrightarrow \mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}),$$ we have $$\begin{align*}(J \cdot (f^{1,0}, 0)) (v \otimes 1) &= (f^{1,0}, 0) (Jv \otimes 1) \\ &= (f^{1,0}, 0) (iv^{1,0} - iv^{1,0}) \\ &= if^{1,0}(v^{1,0}) \\ &= (i \cdot (f^{1,0}, 0))(v \otimes 1), \end{align*}$$ so we have $J (f^{1,0}, 0) = i (f^{1,0}, 0).$ Likewise, we can deduce $J (0, f^{0,1}) = -i (0, f^{0,1})$ for any $$(0, f^{0,1}) \in \mathrm{Hom}_{\mathbb{C}}(V^{0,1}, \mathbb{C}) \hookrightarrow \mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}).$$ This finishes the proof. $\Box$
Note that $x_{1}^{\vee} + iy_{1}^{\vee}, \dots, x_{n} + iy_{n}^{\vee}, x_{1}^{\vee} - iy_{1}^{\vee}, \dots, x_{n} - iy_{n}^{\vee} \in \mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$ form a dual basis for $z_{1}, \dots, z_{n}, \bar{z_{1}}, \dots, \bar{z_{n}} \in V \otimes_{\mathbb{R}} \mathbb{C}.$
Lemma. We have $$x_{1}^{\vee} + iy_{1}^{\vee}, \dots, x_{n} + iy_{n}^{\vee} \in \mathrm{Hom}_{\mathbb{C}}(V^{1,0} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$$ and $$x_{1}^{\vee} - iy_{1}^{\vee}, \dots, x_{n} - iy_{n}^{\vee} \in \mathrm{Hom}_{\mathbb{C}}(V^{0,1} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}).$$ In particular, these elements form $\mathbb{C}$-bases for $\mathrm{Hom}_{\mathbb{C}}(V^{1,0} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$ and $\mathrm{Hom}_{\mathbb{C}}(V^{0,1} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}).$
Proof. Fix a general element $$w = a_{1}z_{1} + \cdots + a_{n}z_{n} + b_{1}\bar{z_{1}} + \cdots + b_{n}\bar{z_{n}} \in V \otimes_{\mathbb{R}} \mathbb{C}.$$ We check $$\begin{align*}(J (x_{j}^{\vee} + iy_{j}^{\vee}))(w) &= (x_{j}^{\vee} + iy_{j}^{\vee})(ia_{j}z_{j}) \\ &= (i(x_{j}^{\vee} + iy_{j}^{\vee}))(a_{j}z_{j}) \\ &= (i(x_{j}^{\vee} + iy_{j}^{\vee}))(w)\end{align*}$$ and $$\begin{align*}(J (x_{j}^{\vee} - iy_{j}^{\vee}))(w) &= (x_{j}^{\vee} + iy_{j}^{\vee})(-ib_{j}\bar{z}_{j}) \\ &= (-i(x_{j}^{\vee} - iy_{j}^{\vee}))(b_{j}\bar{z}_{j}) \\ &= (i(x_{j}^{\vee} - iy_{j}^{\vee}))(w).\end{align*}$$ This implies that $J (x_{j}^{\vee} + iy_{j}^{\vee}) = i (x_{j}^{\vee} + iy_{j}^{\vee})$ so that $$x_{j}^{\vee} + iy_{j}^{\vee} \in \mathrm{Hom}_{\mathbb{C}}(V^{1,0} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$$ due to the previous lemma. Similarly, we get $$x_{j}^{\vee} - iy_{j}^{\vee} \in \mathrm{Hom}_{\mathbb{C}}(V^{0,1} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}),$$ as desired. $\Box$
What's the upshot? Again, we have $$V^{\vee} \otimes_{\mathbb{R}} \mathbb{C} \simeq \mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$$ determined by $v^{\vee} \otimes 1 \mapsto (v \otimes 1)^{\vee},$ which gives $$U^{1,0} \simeq \mathrm{Hom}_{\mathbb{C}}(V^{1,0} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$$ and $$U^{0,1} \simeq \mathrm{Hom}_{\mathbb{C}}(V^{0,1} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$$ the $\mathbb{C}$-bases we found for the right-hand sides produce those of the left-hand sides. Moreover, for any pure tensors $f \otimes a \in V^{\vee} \otimes_{\mathbb{R}} \mathbb{C}$ and $v \otimes b \in V \otimes_{\mathbb{R}} \mathbb{C},$ we can make sense of the evaluation $$(f \otimes a)(v \otimes b) := f(v) \otimes ab \in \mathbb{C},$$ which gives a definition of evaluation of any element of $V \otimes_{\mathbb{R}} \mathbb{C}$ by any element of $V^{\vee} \otimes_{\mathbb{R}} \mathbb{C}.$
Globalization. Let $M$ be a smooth real manifold and $E$ be a smooth real vector bundle on $M.$ An almost complex structure on $E$ is a map $J : E \rightarrow E$ of vector bundles such that $J^{2} = -\mathrm{id}.$ (This makes sense fiber-wise, but it seem okay globally as well.) Note that $J$ will force $E$ have an even rank, say $2r.$ The above discussion globalizes, so we have $$E_{\mathbb{C}} := E \otimes_{\mathbb{R}} \mathbb{C} = E^{1,0} \oplus E^{0,1},$$ coming from fiber-wise description. Everything we discussed in at fiber level works.
What do we mean by tensor product? Write $n = \mathrm{rk}(E)$ for now. We can describe $E$ by various smooth transition functions of the bundle $$(U \cap V) \times \mathbb{R}^{n} \simeq (U \cap V) \times \mathbb{R}^{n},$$ where $U, V$ are from a trivializing chart, which can be presented by smooth maps $U \cap V \rightarrow \mathrm{GL}_{n}(\mathbb{R}).$ We can then consider the smooth maps given by $U \cap V \rightarrow \mathrm{GL}_{n}(\mathbb{R}) \hookrightarrow \mathrm{GL}_{n}(\mathbb{C}),$ which will give us smooth transition functions $$(U \cap V) \times \mathbb{C}^{n} \simeq (U \cap V) \times \mathbb{C}^{n},$$ and these glue back to give $E_{\mathbb{C}}.$
Almost complex structure on tangent bundles. Let $M$ be a smooth real manifold. An almost complex structure on $M$ is an almost complex structure on its tangent bundle $TM$ (i.e., it is a map $J : TM \rightarrow TM$ of bundles with $J^{2} = -\mathrm{id}.$) We write $(TM)_{\mathbb{C}} = T^{1,0}M \oplus T^{0,1}M$ to mean that $T^{1,0}M$ is where $J$ acts as the multiplication by $i,$ while $T^{0,1}M$ by $-i.$ (We will use this notation for other vector bundles and vector spaces from now on.) Note that if $M$ is a complex manifold, then $TM,$ as a real bundle, is already equipped with the complex structure given by multiplication of $i.$
Proposition. Let $M$ be a complex manifold of dimension $n.$ Then there is an almost complex structure on $M$ (or $TM$).
Proof. It is enough to treat the case of (nonempty) open subsets of $\mathbb{C}^{n}$ and then show that biholomorphic maps preserve this structure. Fix any nonempty open $U \subset \mathbb{C}^{n},$ with complex coordinates $z_{j} = x_{j} + i y_{j}$ with $1 \leq j \leq n.$ Then $TU$ has frame (i.e., global sections $U \rightarrow TU$ giving a real basis at each fiber) given by $$\frac{\partial}{\partial x_{1}}, \dots, \frac{\partial}{\partial x_{n}}, \frac{\partial}{\partial y_{1}}, \dots, \frac{\partial}{\partial y_{n}},$$ which gives a trivialization. Define $J : TU \rightarrow TU$ by $$\frac{\partial}{\partial x_{j}} \mapsto \frac{\partial}{\partial y_{j}}$$ and $$\frac{\partial}{\partial y_{j}} \mapsto -\frac{\partial}{\partial x_{j}}.$$ Since $J^{2} = -\mathrm{id},$ this $J$ is an almost complex structure on $TU.$ Hence, we get a decomposition $$TU = T^{1,0}U \oplus T^{0,1}U,$$ where the first component is trivialized by the frame $$\frac{\partial}{\partial z_{1}}, \dots, \frac{\partial}{\partial z_{n}},$$ while the second component is trivialized by the frame $$\frac{\partial}{\partial \bar{z_{1}}}, \dots, \frac{\partial}{\partial \bar{z_{n}}}.$$ Let $V \subset \mathbb{C}^{n}$ be another nonempty open subset. In an earlier lecture, we showed that given any holomorphic map $\phi : U \rightarrow V,$ the map $(TU)_{\mathbb{C}} \rightarrow (\phi^{*}TV)_{\mathbb{C}}$ (which fiber-wise looks like $T_{x}U \otimes_{\mathbb{R}} {\mathbb{C}} \rightarrow T_{\phi(x)}V \otimes_{\mathbb{R}} {\mathbb{C}}$) restricts to $$T^{1,0}U \rightarrow \phi^{*}T^{1,0}V$$ and $$T^{0,1}U \rightarrow \phi^{*}T^{0,1}V.$$ Recall that this actually uses the fact that $\phi$ is holomorphic. In particular, if $\phi$ is biholomorphic, then all the maps above will be isomorphisms of vector bundles preserving $\mathbb{C}$-linearity fiber-wise. $\Box$
Remark. The almost complex structure given on a complex manifold in the proof above is said to be canonical.
Corollary. Let $M$ be a complex manifold of dimension $n.$ Give $M$ the canonical almost complex structure so that we can have the decomposition $$(TM)_{\mathbb{C}} = T^{1,0}M \oplus T^{0,1}M.$$ Then $T^{1,0}M$ is a holomorphic vector bundle.
Proof. It is enough to show that given two nonempty subsets $U, V \subset \mathbb{C}^{n}$ and a holomorphic map $\phi : U \rightarrow V,$ the induced map $T^{1,0}U \rightarrow \phi^{*}T^{1,0}V$ is holomorphic. The induce map $T^{1,0}U \rightarrow T^{1,0}V$ is given by the matrix $$\left[ \frac{\partial \phi_{j}}{\partial z_{k}} \right]_{1 \leq j,k \leq n},$$ and thus each entry is holomorphic. $\Box$
No comments:
Post a Comment