A Tensor Product Identification

130 Views Asked by At

Hi folks just looking at Timmermann's "Introduction to Quantum Groups and Duality" and looking at the algebra of representative functions on a compact topological group $G$.

I take off in Example 3.1.5 (with some slightly different notation)

Let $\rho$ be a continuous representation of $G$ on (a finite dimensional vector space) $V$. Then for every $v\in V$, the function $\chi(v):G\rightarrow V$ given by $x\mapsto \rho(x)v$ belongs to $C(G;V)$. We identify $C(G;V)$ with $V\otimes C(G)$ and and regard $\chi(v)$ as an element of $V\otimes C(G)$.

Presumably $C(G;V)$ is the set of continuous maps from $G$ to $V$. From this remark Timmermann writes

$$\rho(v)=\sum_i e_i\otimes f_i$$

where $\{e_i\}$ is a basis of $V$ and the $f_i\in C(G)$... oh I think I see it now.

Let $\rho(v)(g)=\sum_ia_ie_i$ and suppose $f_i(g)=a_i$. Then

$$\rho(v)\equiv \sum_ie_i\otimes f_i$$

via the identification $V\otimes\mathbb{C}\cong V$.

Is this correct? Sorry these tensor products have me in knots --- pun not intended.

1

There are 1 best solutions below

2
On BEST ANSWER

I think your $C(G)$ denotes here the "dual" $G^\vee=\textrm{Hom}\,(G,\mathbb C)$, the vector space of continuous functions $f:G\to \mathbb C$. There is then an isomorphism of vector spaces $$V\otimes G^\vee\cong\textrm{Hom}\,(G,V).$$ You can see this on generators, and then extend it linearly, by sending $$v\otimes f\mapsto (G\ni x\mapsto f(x)v\in V).$$