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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: 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.2012-01-25

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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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    @Matt Certainly. At least, there should be a common field containing both.2012-01-25
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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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    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.2012-01-25