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Let $E$ and $F$ be two finite dimensional vector spaces. For every $k\in \mathbb{N}$, $E^{k}$ has a natural vector space structure and is isomorphic to $E\otimes \mathbb{R}^{k}$, in a natural way.

We denote by $L^{k}(E)$ $\;$($\Lambda ^{k}(E^{*})$), the space of all $k$-linear maps (anti-symmetric $k$-linear maps) from $E^{k}$ to $\mathbb{R}$. We say that a linear map $T:E^{k}\to F^{k}$ is a tensorial map(anti symmetric tensorial map) if $\forall \alpha \in L^{k}(F)\;(\alpha \in \Lambda^{k}(F^{*}))$, $\alpha \circ T$ belongs to $L^{k}(E)(\Lambda^{k}(E^{*}))$. Clearly such maps form a semi group (with composition) when $E=F$.

A Multilinear question: What is a reference for classification of (the semi group of) all tensorial maps and anti symmetric tensorial maps? Putting $E=F$, is it true to say that an antisymmetric tensorial map is necessarilly in the form $T\otimes S$ for some linear maps $T:E \to E$ and $S:\mathbb{R}^{k} \to \mathbb{R}^{k}$?

Now assume that $E_{1},\ldots E_{k}$, $F_{1},\ldots F_{k}$ are finite dimensional vector spaces. A tensorial linear map $T:\prod E_{i}\to \prod F_{i}$ has a relevant definition.(simillar to the above definition). This definition enable us to ask the following non linear question:

A nonlinear question: Put $M=\mathbb{T}^{n}$. What is an example of an smooth non linear map $f$ on the $n$-torus such that $Df_{x}:T_{x}M\to T_{f(x)} M$ is a tensorial map for all $x\in \mathbb{T}^{n}$? By non linear map We mean" $f$ is not equal to the restriction of a linear map on $\mathbb{R}^{2n}$ which send $\mathbb{T}^{n}$ to $\mathbb{T}^{n}$. What obstruction would appear in non linear case?

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  • $\begingroup$ @HassanJolany Mr. Jolany thanks for your revision. Befor that I read and realize your tag-revision, I was removing the last part of the question and post it in a a new question mathoverflow.net/questions/165755/… $\endgroup$ – Ali Taghavi May 10 '14 at 16:03
  • $\begingroup$ @HassanJolany I am sorry for that. $\endgroup$ – Ali Taghavi May 10 '14 at 16:29
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I don't know whether such a classification exists, but the answer to the second multilinear question is "no":

Consider the case $E=F=\mathbb{R}^2$, $k=3$, and the map $T:(a,b,a',b',a'',b'')\mapsto (a,b,a,b,b',b'')$. This map is linear, but not of tensor form. However, it is antisymmetric tensorial, since $\alpha\circ T=0$ for any $\alpha\in\Lambda^3(F)$.

Concerning the non-linear question: Identify $\mathbb{R}^{2n}\cong\mathbb{C}^n$ and consider the map $f:(z_1,\dots,z_n)\mapsto(z_1^2,\dots,z_n^2)$. Each $D_xf$ is represented by a diagonal matrix, which is tensorial.

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  • $\begingroup$ however your linear example is not a tensorial map. $\endgroup$ – Ali Taghavi May 7 '14 at 21:52
  • $\begingroup$ Yes, but you were asking for an antisymmetric tensorial map. According to your definition, an antisymmetric tensorial map need not be tensorial. It just needs to preserve the set of anti-symmetric k-linear forms. $\endgroup$ – jarauh May 8 '14 at 15:15
  • $\begingroup$ Yes you are right. thanks again for your example. Can I ask you to read the "Motivation" in the last part of my question?may you give some comment on this subject? $\endgroup$ – Ali Taghavi May 8 '14 at 15:25
  • $\begingroup$ I'm sorry, I am not an expert in deRham cohomology. $\endgroup$ – jarauh May 9 '14 at 10:06
  • $\begingroup$ I moved the motivation part to a new question mathoverflow.net/questions/165755/…. $\endgroup$ – Ali Taghavi May 10 '14 at 15:54

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