a question about decomposition of the cotangent space on a complex manifold M (from Griffiths and Harris) Recently, I am reading Griffiths and Harris' book 'principles of algebraic geometry '. In chapter 0, section DeRham and Dolbeault Cohomology. they mention that By linear algebra, the decomposition
$$T^*_{\mathbb{C},z}(M)=T^{*'}_z(M)\bigoplus T^{*''}_z(M)$$
of the cotangent space of a complex manifold M at each point $p\in M$ gives a decomposition
$$\wedge^nT^*_{\mathbb{C},z}(M)=\bigoplus_{p+q=n} (\wedge^p T^{*'}_z(M)\bigotimes \wedge^q T^{*''}_z(M))$$
My question is what is $\bigotimes$ means here? Does it means tensor product? If so, why? In my mind, it should be  $\wedge$ here i.e we have
$$\wedge^nT^*_{\mathbb{C},z}(M)=\bigoplus_{p+q=n} (\wedge^p T^{*'}_z(M)\wedge\wedge^q T^{*''}_z(M))$$
For example, when n=2 (we assume $dim_{\mathbb{C}}M \ge 2$). The decomposition here should be
$$\wedge^2T^*_{\mathbb{C},z}(M)=\wedge^2 T^{*'}_z(M)\bigoplus \wedge^{1,1} T^{*'}_z(M)\bigoplus \wedge^2 T^{*''}_z(M))$$
where
$$\wedge^{1,1} T^{*'}_z(M)=T^{*'}_z(M)\wedge T^{*''}_z(M)=\{f(z)dz_i\wedge dz_{\bar{j}}\} $$
not
$$\wedge^{1,1} T^{*'}_z(M)=T^{*'}_z(M)\bigotimes T^{*''}_z(M)=\{f(z)dz_i\bigotimes dz_{\bar{j}}\} $$
I see many other book use the same symbol as Griffiths and Harris' book (for example 'Complex geometry; an introduction' by Daniel Huybrechts). Can anyone tell what's happening here. Thank you so much.
 A: It's a question of identification.
Suppose $V$ is a (finite dimensional, say) vector space, and $V=U'\oplus U''$ is a decomposition of $V$ as a direct sum of subspaces. The natural map of $\oplus_{k+l =n}  \wedge^k U'\otimes \wedge^l U'' \rightarrow \wedge^n V$, given term-wise by
$$\alpha\otimes \beta \mapsto \alpha \wedge \beta,$$
is an isomorphism, where, on the right of the $\mapsto$, we are identifying  $\alpha$ and $\beta$ (a $U'$ form and $U''$ form respectively) with their images as $V$ forms.
[The map is clearly natural (does not depend on choice of bases, say). But one can use bases (choose a basis for $V$ taken from from bases of $U'$ and $U''$) to prove that the map is an isomorphism.]
The point of the tensor product on the left is that one is considering $U'$ and $U''$ as vector spaces each 'ignorant' of the other (i.e., the $\oplus$ of $U'\oplus U''$ is an external direct sum). The wedge product on the right makes sense for elements of a common [ambient] space. For instance, if $u_1$ and $u_2 \in V$, a popular definition/formula (away from characteristic 2) for $u_1\wedge u_2$ is
$$ u_1\wedge u_2 = 1/2( u_1\otimes u_2 - u_2 \otimes u_1 ).$$
That expression only makes sense because the $u_i$ belong to a common space: one cannot subtract an element of $U''\otimes U'$ from one of $U'\otimes U''$ without some kind of identification.
In your case, $U'$ is the $+i$-eigenspace of $J$, and $U''$ the $-i$-eigenspace of $J$, with $J$ an endomorphism on a vector space $V$ - so both $U'$ and $U''$ are subspaces.
You might like considering (dis?)similar semantics in the situation of an exact sequence of finite dimensional vector spaces
$$0\rightarrow U' \rightarrow V \rightarrow U'' \rightarrow 0.$$
Then there is a isomorphism (not dependent on choice of bases)
$$\wedge^{\rm top} U'\otimes \wedge^{\rm top}U'' \rightarrow \wedge^{\rm top} V,$$
where 'top' are the [top] dimensions of $U', U''$ and $V$ respectively.
And, of course, people 'abuse notation'. For instance,  one might well see things such as $[\alpha_1 \otimes X_1,\alpha_2\otimes X_2]$, or $[(\alpha_1 \otimes X_1)\wedge (\alpha_2\otimes X_2)]$, or ...., where $\alpha_i$ are 'forms,' and $X_i$ elements of a Lie algebra...
