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When working with not-necessarily-finite-dimensional representations, the topology on $GL(V)$ makes a difference. My experience has been that usually people require that the representation $\pi :G\rightarrow GL(V)$ be continuous with respect to the strong-operator topology on $GL(V)$. I would have guessed that instead we would equip $GL(V)$ with the norm topology, so that $GL(V)$ has the nice structure of a Banach manifold. Now that I know this is not the usual assumption, I am guessing this topology is too strong for some purposes. Could someone provide me with some examples demonstrating that the norm topology is the wrong choice of topology when it comes to representation theory?

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$\text{GL}(V)$ isn't a Banach space with the norm topology, it's a Banach manifold (assuming $V$ is a Banach space). –  Qiaochu Yuan Jul 30 '12 at 0:50
    
Oops. I was thinking of $\mathcal{L}(V)$ (i.e., bounded not-necessarily invertible operators). I will edit the post accordingly. –  Jonathan Gleason Jul 30 '12 at 1:06
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See also: mathoverflow.net/questions/66394/… –  t.b. Jul 30 '12 at 5:09

2 Answers 2

up vote 6 down vote accepted

Take the regular representation $$\mathbb{R} \ni t \mapsto (f(x) \mapsto f(x+t))$$

of $\mathbb{R}$ on $L^2(\mathbb{R})$. This is strongly continuous but not norm-continuous, the problem being that the norm of $f(x + t) - f(x)$ can be large even if $t$ is small. But this is clearly a representation we want. (More generally we want the regular representation of $G$ on $L^2(G)$ for $G$, say, a locally compact Hausdorff group.)

Note that if $H$ is a Hilbert space then a norm-continuous one-parameter family of unitary maps $U_t : H \to H$ is necessarily generated by a bounded self-adjoint operator $A$ in the sense that $U_t = \exp (iAt)$ (proof). In the above example the relevant self-adjoint operator is $-i \frac{d}{dx}$, which is not bounded. In general see Stone's theorem.

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It's worth noting that the problem isn't that $\mathbb{R}$ is not compact; this is already an issue for the regular representation of $S^1$ for the same reason, which is about as innocuous an infinite-dimensional representation as you could hope for. –  Qiaochu Yuan Jul 30 '12 at 1:21

Also, strong continuity of $\pi$ is equivalent to continuity of $(g,v)\mapsto \pi(g)v$ (see for example page 23 here: http://www.math.harvard.edu/~jbland/ma222_notes.pdf).

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