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
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