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Suppose $\theta$ is a one-dimensional representation of a group $G$, and $\rho : G \to \mathrm{GL}(V)$ is another representation. Define $\theta \otimes \rho : G \to \mathrm{GL}(V)$ given by $\theta \otimes \rho (g) = \theta(g)\cdot\rho(g)$.

What exactly does the "$\cdot$" mean in the definition of this new map?

Let the vector spaces be over a field $k$. If I interpet $\theta$ as a map $\theta: G \to k^\times$, $\theta(g)$ would be an element of $k$ and so the "$\cdot$" would make sense as scalar multiplication in the vector space $V$. But if $\theta : G \to \mathrm{GL}(W)$ more generally for a one-dimensional vector space $W$, all I can say is that I may write $\theta = \phi \theta' \phi^{-1}$ where $\phi : k \to W$ is an isomorphism and $\theta' : G \to k^\times$. So then $\theta \otimes \rho(g) = \phi \theta'(g) \phi^{-1} \cdot \rho(g)$, and I don't know how to make sense of the "$\cdot$".

I've a feeling I'm being stupid here - any explanation would be most appreciated.

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Dear Matt: You more or less said it: If $\dim V=1$, then $GL(V)$ is canonically isomorphic to $k^\times$. So "$\cdot$" is multiplication in $k$. –  Pierre-Yves Gaillard Jan 24 '12 at 23:59
    
Thanks - but how do I reconcile this with the statement $\theta \otimes \rho(g) = \phi \theta'(g) \phi^{-1} \cdot \rho(g)$? This equivalence seems to make no sense - what am I misunderstanding? –  Matt Jan 25 '12 at 0:06
    
Dear Matt: I suspect that things are clear for you now. Please feel free to tell me if you think I can be of any further help. –  Pierre-Yves Gaillard Jan 25 '12 at 0:43
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2 Answers

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If I understand correctly, your issue is that you think that identification of $\rm{GL}(W)$ with $k^\times$ depends on the choice of basis vector on $W$. But that's clearly not the case: the factor by which an automorphism stretches the 1-dimensional vector space is independent of the basis. To put it differently: the conjugation you write down is always trivial, since $\rm{GL}_1(k)$ is commutative.

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So am I correct in saying that the definition of $\theta \otimes \rho$ doesn't make sense unless $V$ is a vector space over $k$ (i.e. the same field whose units I'm identifying with $\mathrm{GL}(W)$)? [without defining some other multiplication on $V$] –  Matt Jan 25 '12 at 0:16
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@Matt Certainly. At least, there should be a common field containing both. –  Alex B. Jan 25 '12 at 0:33
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If $\theta$ is one-dimensional then it's a map to the $1\times1$ matrices over some field, which matrices can be identified with field elements, so it is just scalar multiplication in the vector space $V$. I think you're confusing yourself when you write $GL(W)$ for a vector space $W$; don't you mean $GL(F)$ for the field $F$?

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Thanks for your answer. By $\mathrm{GL}(W)$ I mean the group of invertible linear maps $W \to W$; this is the notation my notes use. I suppose I mean $\mathrm{GL}_F(V)$ more precisely. Have I been silly when assuming $V$ and $W$ above are over the same field $k$? If so, I'm again confused. –  Matt Jan 25 '12 at 0:10
    
I think we were both confused. I think you can take it as implied that the two representations mentioned in your quote are over the same field. –  Gerry Myerson Jan 25 '12 at 1:40
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