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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.

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Ali Taghavi
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  • What is a $k$-tensor? Doesn't the whole group $GL$ act on the whole tensor algebra? What is this special "tensorial" condition? – Alex Degtyarev May 10 '14 at 15:57
  • @AlexDegtyarev by $k-tensor$ I mean a k-linear map from $E^{k}$ to $\mathbb{C}$. – Ali Taghavi May 10 '14 at 16:08
  • @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. – Ali Taghavi May 10 '14 at 16:12
  • @AlexDegtyarev is my question clear, now? – Ali Taghavi May 10 '14 at 18:56
  • Yes, it's clear, but I'm not exited. Concerning your last question, as I said, any subgroup of $GL(E)$ (e.g., $O(E)$ if you want it compact) would act on any tensor power. As to the atlas, you're speaking about the "structure group reduction", which is a well developed subject. – Alex Degtyarev May 10 '14 at 20:18
  • @AlexDegtyarev I Think there is a misunderstanding on the notion of action: are you saying that for every $k$-linear map $\alpha:E^{k} \to \mathbb{C}$ and every arbitrary linear map $T\in GL(E^{k}$, $\alpha \circ T$ is necessarily a $k$-linear map?(I think it is not true). Note that my action was simply the "composition". What action you are talking about? – Ali Taghavi May 10 '14 at 21:12
  • Unlike you, I'm speaking about $GL(E)$, not $GL(E^k)$. All I know is that $T^*$ is a functor (as so are $\otimes$ and $\operatorname{Hom}$), and the action is obvious. – Alex Degtyarev May 10 '14 at 21:15
  • Thank you. so you introduce the product $G=\prod_{i=1}^{k} O(E)$ as an example which I was searching for. So we choose "Determinant" as a real valued character(1,-1). Does this situation leads to a nontrivial cohomology? What would be a (differential) operator $\tilde{d}$? Is "determinant" the only possible character? – Ali Taghavi May 10 '14 at 21:30
  • Your G is contained in one G_k, and you arenviewing it as acting on the higher than kth tensors by the inclusion of G_k into G_{k+1} that you mention. First, this does not tell us how G acts on lower tensors, and second if you do this with S_k you certainly do not get the usual de Rham complex! – Mariano Suárez-Álvarez May 11 '14 at 21:05
  • With $G=\prod O(E)$ as in your last comment, the wedge of any two vectors is zero, since the average of any vector over $SO(E)$ is zero. This sort of kills any chance of defining $\tilde d$ in the usual way, by its action on 0-forms and then extending by wedge products. Also, I'm not sure your definition of invariant extends the classical case-- for example $e_1\wedge e_2$ is fixed by $(3,4)$, hence not invariant in your sense. I think you mean the basis elements of $\Lambda^k$ should be $k$-fold wedge products. – Brendan Murphy May 11 '14 at 21:36
  • @BrendanMurphy Thanks for your comment. First, I work with covectors not vectors. second put $E=\mathbb{R}$. $k=2$. $G=O(\mathbb{R})\times O(\mathbb{R})$, the character $\phi=Det$. Then we have $dx\tilde{\wedge} dx=4 dx\otimes dx$, a non zero tensor. In fact I am introducing a new wedge product. For more detail, please read my comment as a response to Mariano's comment(Which I will wrote up to 30 minute. – Ali Taghavi May 12 '14 at 21:11
  • @MarianoSuárez-Alvarez thank you for your attention on my post. I think there is an interesting case which my approach works, well. Put $E=\mathbb{R}^{2n}$. Identify $E$ with $\mathbb{C}^{n}$. Let $H$ be a torus in $GL(E)$ with the property that for every holomorphic map $f: U \to \mathbb{C}^n$, $U\subset \mathbb{C}^n$, all $T\in H$ commutes with $J(f)_{x}$ as a real linear map on $\mathbb{R}^{2n}$. Ex: for $n=1$ this Torus is the group of rotations.(I do not know what is the maximal dimension for such torus?). Choose and fix a character $\phi$ for $H$. I continue in the next comment – Ali Taghavi May 12 '14 at 21:27
  • My two previous comments can not be read, well. i can not delete or revised them. so i summarize both comments in the following. @BrendanMurphy Put $E=\mathbb{R}$, $k=2$ $G=\prod O(\mathbb{R})$. Then $dx \tilde{\wedge} dx =4 dx \otimes dx$, a non zero tensor. – Ali Taghavi May 12 '14 at 21:44
  • @AliTaghavi I added some dollar signs to your comments. – S. Carnahan May 12 '14 at 23:21
  • @S.Carnahan Thank you very much for your help. – Ali Taghavi May 12 '14 at 23:24
  • @AliTaghavi Well in the case $E=R^1$, $SO(E)$ is the identity group, but in the case $E=R^2$, $dx\tilde\wedge dx$ is composed of four terms like $\int_{\rho_1\in SO(2)}\int_{\rho_2\in SO(2)}d(\rho_1 x)\otimes d(\rho_2 x)$, and each integral is zero since a vector gets averaged over $SO(2)$ before it gets plugged in to $dx$. – Brendan Murphy May 13 '14 at 00:21
  • @BrendanMurphy Thanks for your interesting comment. I think you use 'the separation variable" in your computation.Yes?This computation is valid because of the special form of $G$:it is a diagonal product. This is a motivation to consider some groups which is not in the product form(to avoid the triviality). however in the case you considered(with $G=SO(E) $ not $O(E)$, we have $\alpha \tilde{\wedge} \alpha \neq 0$ when $\alpha= dx \otimes dx$. So it is not true that our (averaging) wedge product is always zero.. – Ali Taghavi May 13 '14 at 12:10
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    @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 $$\alpha\tilde\wedge\beta((x,y),(u,v))=\int_0^{2\pi}\int_0^{2\pi}(x\cos\theta+y\sin\theta)(u\cos\phi+v\sin\phi)d\theta d\phi =0.$$ – Brendan Murphy May 13 '14 at 16:51
  • @BrendanMurphy yes, I see. thank you. – Ali Taghavi May 13 '14 at 17:01
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    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. – Brendan Murphy May 13 '14 at 17:23

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