Inducing almost complex structure in tensor bundle

Question

Let $E\to M$ be a (smooth) vector bundle and $J$ be a section of ${\rm End}(E)$ with $J^2= -{\rm Id}$. Can $J$ induce something in $$\mathscr{T}^{(r,s)}(E)=\bigsqcup_{x\in M} E^{\otimes r}\otimes (E^*)^{\otimes s},$$in general? For example, in $E^* = \mathscr{T}^{(0,1)}(E)$ one can set $(J^*\zeta)(\psi)= \zeta(J\psi)$, this $J^*$ squares to $-{\rm Id}$ and has the nice property that whenever we have a connection $\nabla$ in $E$, the identity $(\nabla_XJ^*)(\zeta)= \zeta\circ \nabla_XJ$ holds. But if I try to combine these two in the general case and set $$(J\Phi)(\zeta^1,\ldots,\zeta^r,\psi_1,\ldots,\psi_s) = \Phi(\zeta^1\circ J,\ldots ,\zeta^r\circ J,J\psi_1,\ldots, J\psi_s),$$I get the awkward sign $J^2 = (-1)^{r+s} {\rm Id}$ (which at least is consistent what with I did for $E^*$).

To summarize my question: I don't like this sign. I wanted $J^2=-{\rm Id}$ always. Is this fixable? If not, are there deeper reasons?

Answer 1

Let's consider the linear algebra question. Assume $E,F$ are two finite dimensional real spaces with complex structures $J_1 \colon E \rightarrow E$ and $J_2 \colon F \rightarrow F$. The space $E^{*} \otimes_{\mathbb{R}} F$ can be naturally identified with the space $ \operatorname{Hom}_{\mathbb{R}}(E,F)$ of $\mathbb{R}$-linear maps from $E$ to $F$. We have the following maps:

1. The map $\hat{J}_1 = J_1^{*} \otimes \operatorname{id}_E$ satisfies $\hat{J}_1^2 = (J_1^{*} \circ J_1^{*}) \otimes (\operatorname{id}_F \circ \operatorname{id}_F) = -\operatorname{id}_{E^{*}} \otimes \operatorname{id}_F = -\operatorname{id}_{E^{*} \otimes_{\mathbb{R}} F}$ so it gives us a complex structure on $E^{*} \otimes_{\mathbb{R}} F$. Under the identification of $E^{*} \otimes_{\mathbb{R}} F$ with $\operatorname{Hom}_{\mathbb{R}}(E,F)$, this corresponds to using the complex structure on the domain $E$ to define the complex multiplication on the space of real mappings. Namely, if $T \colon E \rightarrow F$ is a real linear map, we have on $(\operatorname{Hom}_{\mathbb{R}}(E,F), \hat{J}_1)$ $$ (iT) := \hat{J}_1(T) = T \circ J_1, \\ (iT)(v) = T(J_1v) = T(iv). $$
2. The map $\hat{J}_2 := \operatorname{id}_{E^{*}} \otimes J_2$ also satisfies $\hat{J}_2^2 = -\operatorname{id}_{E^{*}} \otimes \operatorname{id}_F$ so it gives us another complex structure on $E^{*} \otimes F$. Under the identification of $E^{*} \otimes_{\mathbb{R}} F$ with $\operatorname{Hom}_{\mathbb{R}}(E,F)$, this corresponds to using the complex structure on the range $F$ to define the complex multiplication on the space of real mappings. Namely, if $T \colon E \rightarrow F$ is a real linear map, we have on $(\operatorname{Hom}_{\mathbb{R}}(E,F), \hat{J}_2)$ $$ (iT) := \hat{J}_2(T) = J_2 \circ T, \\ (iT)(v) = J_2(Tv) = iT(v). $$
3. The map $\hat{J} = J_1^{*} \otimes J_2$ which corresponds to your suggestion. Indeed, $$\hat{J}^2 = (-\operatorname{id}_{E^{*}}) \otimes (-\operatorname{id}_F) = \operatorname{id}_{E^{*}} \otimes \operatorname{id}_F$$ so it is not a complex structure. Under the identification of $E^{*} \otimes_{\mathbb{R}} F$ with $\operatorname{Hom}_{\mathbb{R}}(E,F)$, this corresponds to using the complex structure both on the domain and range $F$ to define an action on the space of real mappings. Namely, if $T \colon E \rightarrow F$ is a real linear map, we have $$ \hat{J}(T) = J_2 \circ T \circ J_1, (\hat{J}(T))(v) = iT(iv). $$ The map $\hat{J}$ has eigenvalues $\pm 1$ with eigenspaces $$ \{ \hat{J}(T) = T \} = \{ J_2 \circ T \circ J_1 = T \} = \{ T \circ J_1 = -J_2 \circ T \} = \{ T \colon E \rightarrow F \, | \, T \textrm{ is } \mathbb{C}-\textrm{antilinear} \}, \\ \{ \hat{J}(T) = -T \} = \operatorname{Hom}_{\mathbb{C}}(E,F). $$ Using $\hat{J}$, we can decompose any real linear map $T \colon E \rightarrow F$ into its $\mathbb{C}$-linear and $\mathbb{C}$-antilinear parts as $$ T = \frac{1}{2}\left( T - \hat{J}(T) \right) + \frac{1}{2} \left( T + \hat{J}(T) \right) = \frac{1}{2} \left( T - J_2 \circ T \circ J_1 \right) + \frac{1}{2} \left( T + J_2 \circ T \circ J_1 \right). $$

Note that we have $\hat{J} = \hat{J}_1 \circ \hat{J}_2 = \hat{J}_2 \circ \hat{J}_1$ so $\hat{J}$ is a product of two commuting complex structures. Also note that even if $E = F$ and $J_1 = J_2$ (which is your case), the complex structures $\hat{J}_1$ and $\hat{J}_2$ are different. Of course you can also replace $J_1,J_2$ with $-J_1,-J_2$ and get the conjugate complex structures.

In general, you can put $r+s$ different complex structures on $(E^{*})^{\otimes s} \otimes E^{\otimes r}$ which pairwise commute and induce various direct sum decompositions. Multiplying any odd number of them will give you additional complex structures.