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Interpret the representation $ T \otimes T$ in terms of matrices, and compare it with $T^2$.

Could anyone give me a hint on how to solve this please ?

EDIT:

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Answer with correct definitions:

First of all let me clearly state the definitions involved:

Let $T,S$ be $G$-representations on vector spaces $U$ and $V$ respectively. Then the product $T\cdot S$ is a $G$-representation given by $(T\cdot S)(g)=T(g)\otimes S(g)$.

Let $T$ be a $G$ representation on $U$ and let $S$ be an $H$-representation on $V$. The tensor product $T\otimes S$ is a $G\times H$-representation on $U\otimes V$.

Now let $T$ be a $G$-representation on $U$. To be precise, we have $T\cdot T:G\rightarrow \text{GL}(U\otimes U):g\mapsto T(g)\otimes T(g)$ and $T\otimes T:G\times G\rightarrow \text{GL}(U\otimes U): (g,h)\mapsto T(g)\otimes T(h)$. Hence a first major difference is that these are representations of different groups!

Now let $\left\{e_1, \dots , e_n\right\}$ be a basis of $U$. Any element $x\in U\otimes U$ can be uniquely expressed in the form $x=\sum_{i,j=1}^nx_{ij}(e_i\otimes e_j)$ with $x_{ij}$ in the field you are working over (usually $\mathbb{R}$ or even better $\mathbb{C})$. Denote by $X$ the matrix $[x_{ij}]$.

Now we can ask ourselves two questions: How is $X$ transformed under $T^2(g)$ and under $(T\otimes T)(g,h)$?

Convince yourself of the following: $T^2(g)(X)=T(g)XT^*(g)$ where $T^*$ is the dual (or contragredient) representations of $T$ and $(T\otimes T)(g,h)(X)=T(g)XT^*(h)$.

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  • $\begingroup$ So what you are saying that the main difference is that the dual operator acts on an element of the group G in the product while in the tensor product it acts on an element of the group H ..... am I correct? $\endgroup$ – Intuition Feb 7 at 14:47
  • $\begingroup$ Can you give me a concrete ( numeric )example of doing the tensor product $T \otimes T$ and T^2 as I am a little bit confused from those 2 definitions $\endgroup$ – Intuition Feb 7 at 16:27
  • $\begingroup$ As for your first comment: No, that's not what I'm saying. You start with only a $G$-representation $T$. As far as this question is concerned, there is no group $H$ involved. The main difference is that $T^2$ is a $G$-representation and $T\otimes T$ is a $G\times G$-representation. $\endgroup$ – Mathematician 42 Feb 7 at 19:19
  • $\begingroup$ And also the tensor product must act on different elements even if we are in the same group while the product acts on the same element .... am I correct? $\endgroup$ – Intuition Feb 7 at 21:10
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    $\begingroup$ I'm not willing to write out a specific example. Having said that, take for example the symmetric group $S_3$ and let $T$ be a two-dimensional representation. Follow the steps outlined above and compare. $\endgroup$ – Mathematician 42 Feb 8 at 8:20
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You are missing a lot of details. Anyway, let $T:V\rightarrow W$ be a linear map between finite dimensional vector spaces over some field $k$ and let $\dim(V)=n$ and $\dim(W)=m$. After a choice of bases, the map $T$ can be represented by a $m\times n$ matrix over $k$.

By definition, $T\otimes T: V\otimes V\rightarrow W\otimes W$ and thus can be represented by a $m^2\times n^2$ matrix over $k$. On the other hand, $T^{2}=T\oplus T:V\oplus V\rightarrow W\oplus W$ can be represented by a $2m\times 2n$ matrix over $k$.

So in general, $T\otimes T$ and $T^2$ are not even represented by matrices of the same size. Perhaps this already indicates how different both maps are!

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  • $\begingroup$ What do you mean by I am missing a lot of details $\endgroup$ – Intuition Feb 6 at 14:41
  • $\begingroup$ I will edit the question to include my book definitions of product and tensor product of representations .... I am working from Vinberg " linear representations of groups" $\endgroup$ – Intuition Feb 6 at 17:22
  • $\begingroup$ I'm confused why you accepted my answer. Based on the definitions you now included with links (which you shouldn't really do on this site), my answer doesn't apply to this situation. In fact, my interpretation of $T$ as a linear operator and my interpretation of $T^2$ as a direct sum of operators is not what is intended by your course! $\endgroup$ – Mathematician 42 Feb 7 at 8:28
  • $\begingroup$ I liked the kind of thought in your first answer .... but sure I will accept the second one :) $\endgroup$ – Intuition Feb 7 at 14:45

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