# Is this a correct interpretation of support in coarse geometry?

Let $X = \mathbb{R}^n$, and consider a nondegenerate representation $\rho: C_0(X) \to B(H)$ where $B(H)$ is the algebra of bounded operators on a separable Hilbert space. The support of a vector $v \in H$ is defined to be the complement in $X$ of the union of all open sets $U$ such that $\rho(f)v = 0$ for every $f \in C_0(U)$.

Suppose $v$ has compact support. My intuition is that any function $g \in C_0(X)$ which restricts to the zero function on $supp(v)$ should satisfy $\rho(g)v = 0$, but I can't quite prove it. Here is what I have so far.

Let $\mathcal{F}$ denote the collection of open sets $U$ such that $\rho(f)v = 0$ for $f \in C_0(U)$. By definition $V = supp(v)^c$ is the union of the open sets in $\mathcal{F}$, and moreover a simple partition of unity argument shows that $\mathcal{F}$ is closed under finite unions. If we can show that $V \in \mathcal{F}$ then we are done because by the hypotheses $g \in C_0(V)$. I'm sure I'm just missing something simple; can anyone help?

This result should be true if $X$ is any separable metric space equipped with a proper coarse structure, so I suppose the proper setting for this question is metric geometry. That should explain the title of the question and the tags.

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I don't know anything about coarse geometry, but it seems prudent to indicate that the question is about commutative Banach algebras, either via tags or via title. – Victor Protsak Jun 13 '10 at 7:22
By "non-degenerate" do you mean that \rho is a *-homomorphism, with $\{\rho(f)v : f\in C_0(X), v\in H\}$ linearly dense in H? – Matthew Daws Jun 13 '10 at 8:46
There is a small complication that while $g=0$ outside of $V,$ this doesn't imply that $g\in C_0(V).$ My first proof assumed $g\in C_0(V)$ and following Matthew's suggestion, I did the general case. – Victor Protsak Jun 13 '10 at 11:33
That was probably a good tagging decision - thanks. – Paul Siegel Jun 13 '10 at 14:50
@Mathew: that is indeed the notion of nondegenerate I am using here. Though for this problem I suppose it may not matter. – Paul Siegel Jun 13 '10 at 14:51

Edit I have amended the proof to cover the general case following a suggestion of Matthew Daws.

By the definition of $supp(v)$, for any $x$ in $supp(v)^c$ there exists an open set $U(x)\subset supp(v)^c$ containing $x$ such that $\rho(f)v=0$ for all $f\in C_0(U(x)).$ If $g$ has compact support $K\subset supp(v)^c,$ there is a finite subset $U_1,\ldots,U_m$ of $\{U(x)\}$ covering $K.$ Using a partition of unity, $g=g_1+\ldots+g_M$ where $g_i\in C_0(U_{k(i)})$. Therefore, $\rho(g)v=\sum_i \rho(g_i)v=0.$ In general, by the lemma below, we can approximate $g$ by a sequence $\{g_n\}$ of continuous functions with compact support disjoint from $supp(v),$ so $\rho(g)v=\rho(\lim g_n)v=\lim \rho(g_n)v=0.\square$

Lemma Suppose that $L\subset X$ is compact and $g\in C_0(X)$ restricts to zero function on $L.$ Then $g=\lim g_n,$ where $g_n\in C_0(X)$ has compact support disjoint from $L.$

Proof The function $g_n$ is obtained from $g$ by a smooth cutoff at distance $1/n$ from $L.$ The approximation property follows from the fact that $g$ vanishes on $L$.

More formally, let $h:\mathbb{R}\to [0,1]$ be a continuous function such that $h(y)=0$ for $y\leq 1$, $h(y)=1$ for $y\geq 2.$

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/     h(x)
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1   2


Since $L$ is compact, the distance function $d_X(\cdot,L)$ is well-defined and continuous. Let $L_n$ be the open $1/n$-neighborhood of $L$ in $X$. Set

$$g_n(x)=h(nd_X(x,L))g(x).$$

By construction, $g_n$ vanishes on $L_n$ and coincides with $g$ on $L_{2n}^c$. Its support is compact and is contained in $L_n^c.$ Moreover,

$$\|g-g_n\|\leq \sup_{x\in {L}_{2n}} |g(x)|.$$

The right hand side is a non-negative monotone decreasing sequence. Suppose that there exists a sequence of points $x_n\in L_{2n}$ such that $g(x_n)$ is bounded away from $0.$ Since $g$ is compactly supported, this sequence has an accumulation point $x.$ Then $x\in L$ and so $g(x)=0,$ which is a contradiction. $\square$

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If $L$ is compact, and $g\in C_0(X)$ vanishes on $L$, it is possible to approximate $g$ by $f$, where $f$ has compact support disjoint from $L$? Intuition suggests "yes", but I'm not sure... – Matthew Daws Jun 13 '10 at 8:51
Yes. Let $L_n$ be the open $1/n$-neighborhood of $L.$ Then $g$ can be approximated by $\{f_n\}$ with compact support in $L_n^c$ using a rescaled bump function composed with $d(\cdot,L).$ – Victor Protsak Jun 13 '10 at 9:50
So that would complete the proof right? If g vanishes on $supp(v)$ then approximate g by $(f_n)$ which all have compact support in $supp(v)^c$. Then $\rho(g)v = \lim_n \rho(f_n)v = 0$. Or did I miss something? – Matthew Daws Jun 13 '10 at 10:28
Yes, and I have completed it in the meantime (amusingly, I didn't see your comment until now, but I've given you proper credit nonetheless :) – Victor Protsak Jun 13 '10 at 11:27
Thank you both for your help! This looks like it works perfectly, and as I hoped it generalizes immediately to the class of metric spaces that I need. – Paul Siegel Jun 13 '10 at 14:48

If $\rho$ is a *-homomorphism, then I'd be tempted to use a little bit of C*-algebra theory. Pick a maximal family $\{v_i\}$ of unit vectors in $H$ such that $H_i = \overline{\operatorname{lin}}\{ \rho(f)v_i : f\in C_0(X) \}$ are mutually orthogonal. Then there is a probability measure $\mu_i$ on X such that $(\rho(f)v_i|v_i) = \mu_i(f)$ for $f\in C_0(X)$, and each $H_i$ is unitarily equivalent to $L^2(X,\mu_i)$, with $\rho$ being transformed to the canonical action of $C_0(X)$ on $L^2(X,\mu_i)$.

That is, H is just the direct sum of spaces $L^2(X,\mu)$. So, for the moment, let's just suppose that H is $L^2(X,\mu)$. I'm a touch worried that the condition $f\in C_0(U)$ is a little bit weaker than $f$ having support contained in $U$, but modulo some details, surely the definition of "support" in the original post is the same as the usual definition of support for a measure. If then $g\in C_0(X)$ vanishes on the support, then immediately $\int_X |g| d\mu = 0$ (if one believes Wikipedia).

If $H$ is the direct sum of $L^2(X,\mu)$, and $v=\sum a_i v_i$ say, then if $J=\{i:a_i\not=0\}$ we have that $\rho(f)v=0$ if and only if $\rho(f)v_j=0$ for all $j\in J$. So things get a bit tricky here, as if $J$ is infinite, the support of $v$ is probably the closure of union of the supports of the $\mu_j$, for $j\in J$. But if $g\in C_0(X)$ then vanishes on this, it has to vanish on $supp(\mu_j)$ for all $j\in J$, which is enough to show that $\rho(g)v_j=0$ for all $j$, showing $\rho(g)v=0$. I hope...

But, it seems that this argument isn't easier than Victors...

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