Edit According to the comment of Alex Degtyarev, I deleted the last part of the previous version.

Let $E$ be a real vector space. The complex valued $k$- tensors on $E$ is denoted by $L_{\mathbb{C}}^{k}(E)$, that is, the space of all k-linear maps from $E^{k}$ to $\mathbb{C}$. A linear map $T$ in $GL(E^{k})$ is called a tensorial map if $\forall \alpha \in L_{\mathbb{C}}^{k}(E)$, the composition $\alpha \circ T$ belongs to $L_{\mathbb{C}}^{k}(E)$. For example a permutation in $S_{k}$ gives us a tensorial linear map on $E^{k}$. Let $G_{K}$ be the subgroup of $GL(E^{k})$ consists all tensorial maps in $GL(E^{k})$. There is a natural embedding from $G_{k}$ to $G_{k+1}$ which sends $T$ to $T\oplus Id_{E}$.

Assume that $G$ is a compact topological subgroup of $G_{k}$. (For example, the permutation group $S_{k}$ is embedded in $G_{k}$.

We fix a character $\phi:G \to S^{1}$.(For example $\phi$ is the $\it{sign}$ character, when $G$ is $S_{k}$). We assume that $\mu$ is the Haar measure(ex:Counting measure for $S_{k}$). We say that $\alpha \in L_{\mathbb{C}}^{k}(E)$ is $G$ invariant if $\alpha \circ T=\phi (T) \alpha, \forall T\in G$.(For $G=S_{k}$, this definition, gives us the definition of ordinary "k-forms" on $E$). We denote the collection of all $G$ invariant tensors by $\tilde{\Lambda}(E^{*})$.

For two $G$-invariant tensors $\alpha$ and $\beta$ we define $\alpha \tilde{\wedge} \beta:=\int_{G} \phi(T) (\alpha \otimes \beta)\circ T$, which is a $G$-invariant tensor. This definition is motivated by the ordinary wedge product of ordinary "forms".

A linear map $S$ on $E$ is called $G$-invariant if $\oplus_{i=1}^{k} S$ commutes with all operators in $G$.(When $G=S_{k}$, all linear maps $T$ on $E$, are automatically,$G$-invariant). Obviously the pull back of an invariant tensor, under an invariant linear map, is again an invariant tensor. Now under the following condition on a manifold $M$,we can non-linearize this process to obtain $\tilde{\Omega}^{k}(M)$, in the same manner as the ordinary DeRham cohomology: the condition is that we have an atlas for $M$ such that the jacobian of each transition map is $G$ invariant. Now a natural question is that:

"Is there a natural complex of (differential) operators $\tilde{d}:\tilde{\Omega}^{k}(M) \to \tilde{\Omega}^{k+1}(M)$ ? Is it useful to produce and study such kind of cohomology? Does such cohomology give us new information, different from information which is obtained from deRham cohomology? What is an algebraic topological version of such smooth cohomology?(motivating by the fact that singular cohomology is parallel to the DeRham cohomology). Under what conditions we have an atlas such that the jacobian of transition maps are $G$ invariant?

Note Acording to the comment of Mariano I add the following example:Put $E=\mathbb{R}^{2n}$ which is identified with $\mathbb{C}^{n}$. Let $H$ be a torus in $GL(E)$ which all operators, commute with the real-Jacobian of all local holomorphic map from open sets of $\mathbb{C}^{n}\to \mathbb{C}^{n}$. Ex; $n=1$ $H=$ rotations.

Then $H^{k}=\prod_{i=1}^{k} H$ is a subgroup of $G_{k}$ and the natural embedding of $G_{k}$ into $G_{k+1}$is compatible with the natural embedding of $H^{k}$ into $H^{k+1}$ which add the identity element. Choose and fix a character $\phi$ on $H$. Then define the character $\phi_{k}$ on $H^{k}$ with $\phi_{k}=\prod \phi$. So $\phi_{k+1}$ is an extension of $\phi_{k}$. So I think the approach of this post is applicable to holomorphic manifolds, provided we choose and fix in priori, $H$ and $\phi$ as above.

  • $\begingroup$ What is a $k$-tensor? Doesn't the whole group $GL$ act on the whole tensor algebra? What is this special "tensorial" condition? $\endgroup$ – Alex Degtyarev May 10 '14 at 15:57
  • $\begingroup$ @AlexDegtyarev by $k-tensor$ I mean a k-linear map from $E^{k}$ to $\mathbb{C}$. $\endgroup$ – Ali Taghavi May 10 '14 at 16:08
  • $\begingroup$ @AlexDegtyarev Now I realize what you mean. I just consider 0-k tensors(If i am not mistaken in terminology). so $GL(E^{k}$ does not act on it. $\endgroup$ – Ali Taghavi May 10 '14 at 16:12
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    $\begingroup$ @AliTaghavi Yes, the product group lets you use Fubini's theorem to compute the integral. If $G=O(E)$ there are 4 terms instead of 1 because $O(E)$ is a union of two cosets of $SO(E)$. Each integral is zero if $E$ has dimension greater than 1. I will compute one integral explicitly: let $\alpha(x,y)=x$ and let $\beta(x,y)=x$. Then $$\alpha\tilde\wedge\beta((x,y),(u,v))=\int_0^{2\pi}\int_0^{2\pi}(x\cos\theta+y\sin\theta)(u\cos\phi+v\sin\phi)d\theta d\phi =0.$$ $\endgroup$ – Brendan Murphy May 13 '14 at 16:51
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    $\begingroup$ Another way to see this is through the mean ergodic theorem. You are considering unitary operators acting on the Hilbert space of $k$-forms, so averaging over any one parameter group will result in a projection on to the fixed points of that group---namely $L_{\mathbb{C}}^k(E)$. If you have non-trivial action on the individual components, it seems like it will just reduce you to a smaller dimension base space (at least looking at $k=1$); thus like you mentioned it seems that the groups that mix are the interesting ones. $\endgroup$ – Brendan Murphy May 13 '14 at 17:23

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