3
$\begingroup$

It is classical that $L^1(G, m)$ is a Banach algebra when $G$ is a locally compact group with Haar measure $m$ by using the operation of convolution via the integral

$$(f*g)(y)=\int_Xf(x)g(yx^{-1})\,dm(x)$$

I'm interested in knowing if there is a natural Banach algebra structure on $L^1(X, \mathcal{B},\mu)$ for a locally compact, Hausdorff measure space $X$, with a Borel sigma algebra, $\mathcal{B}$ and radon measure, $\mu$ with the added assumption that the measure is induced by a (say normalized) Haar measure on a compact abelian group, $G$.

In particular, the situation I'm most interested in is

$$X=\coprod_{n\in\mathbb{N}} X_n$$

with each $X_n$ a compact space topologically a homogeneous $G$-space, i.e. a quotient space $G/H_n$ with $H_n$ closed subgroups of $G$. With these assumptions I can define just such a measure on $X$ by first taking $f\in C(X_n)$ and selecting any point $x_n\in X_n$ and using the normalized Haar measure $m$ on $G$, setting

$$I_n(F)=\int_G f(\tau x_n)\,dm(\tau)$$

Since $G$ acts transitively on its coset spaces, this integral is independent of choice of $x_n$. Then, on $X$, I define for $F\in C_c(X)$

$$I(F)=\sum_n I_n(F\cdot 1_{X_n})$$

Since $F$ has compact support, this sum is well-defined (being finite). I know that, since $I(F)$ is a linear functional, it is represented by integration against a Radon measure, $\lambda$ on $X$ and by construction and the transitivity on the pieces, this measure is invariant under the action of $G$, that is to say, there is a radon measure $\mu$ on $X$ and a $\sigma$-algebra, $\Sigma$, containing the Borel $\sigma$-algebra so that for every $F\in L^1(X,\Sigma,\mu)$

$$I(F)=\int_X F\,d\mu.$$

Moreover it is clear from the definition that this has the property that

$$\int_{X_n}F(x)\,d\mu(x)=\int_{X_n}F(\tau x)\,d\mu(x)$$

for every $\tau\in G$ and $F\in L^1(X)$. By writing

$$F=\sum_n F\cdot 1_{X_n}$$

we can see now that

$$\int_X F\,d\mu=\sum_n F(x)1_{X_n}\,d\mu(x)$$ $$=\sum_n\int_{X_n}F(x)\,d\mu=\sum_n \int_{X_n}F(\tau x)\,d\mu(x)$$ $$=\int_X F(\tau x)\,d\mu(x).$$

This establishes that the measure $\mu$ is invariant under $G$.

Now, $X$ itself lacks a group structure, but can we still somehow define a Banach algebra structure on $L^1(X)$ by using the fact that the measure $\mu$ is essentially pushing forward the measure $m$ on $G$ and $G$ does have the structure of a compact group?

Edit (Adam): I've added the fact about the $X_n$ being homogeneous spaces of $G$ by subgroups and clarified the invariance of $\mu$ under the $G$-action.

$\endgroup$
  • $\begingroup$ Since you are working with an homogeneus $G$-space $X$ the action of $G$ in $X$ will not induce an algebra structure on $L_1(X)$ but a natural Banach $L_1(G)$-module structure (right or left depending on your choice of actions). I do not know if that is a satisfactory answer. $\endgroup$ – Adrián González-Pérez Nov 28 '13 at 7:08
2
$\begingroup$

If the $H$ are compact and $G$ unimodular, there is indeed a algebra structure on $L^1(G/ H)$ simply by convolution over $G$. It does however not contain more structure then the subalgebra $A=L_1( G//H) \subset L^1(G)$ of $H_n$-biinvariant functions.

You can drop modularity of $G$ an will obtain a slightly different $H$-invariance from the left.

This actually works fine still if $H$ is containing and compact modulo the center $Z$ of $G$. You obtain as subalgebras $$L^1(G//H) \subset L^1(G/Z).$$

This naive approach fails if $G//H$ is not nicely defined, say $H$ being the subgroup of upper triangular matrices in $G=GL_2(\mathbb{R})$. Then $G//H$ has cardinality two, and one of the double coset carries the full measure, the other zero measure.

This does of course not rule out any different Banach algebra structure on $L^1(G/H)$, but $G/H$ is pretty well-known (i.e. projective space). So perhaps you should start with that specific example?

$\endgroup$
  • $\begingroup$ It seems you've addressed the compact case. For the situation described, does one just do the natural thing and define the action of convolution on $X$ by writing integrals as sums over the component pieces to get the structure on a union of such $G$-spaces then, or is there some subtlety I'm potentially not seeing? $\endgroup$ – Adam Hughes Nov 29 '13 at 9:49
  • $\begingroup$ Yes, you can do the usual integration on $G$ and ignore invariance. $\endgroup$ – Marc Palm Nov 29 '13 at 14:40
1
$\begingroup$

I am not sure that this will be useful for you, but Yulia Kuznetsova proved not long ago that for a locally compact group $G$ and a measurable function $w:G\to{\mathbb R}_+$ the weighted space $$ L_1^w(G)=\{f:\quad fw\in L_1(G)\} $$ is an algebra (with respect to the convolution) if and only if the weight $w$ is equivalent in some sense to a continuous submultiplicative function: $$ w(st)\le w(s)w(t), \qquad s,t\in G. $$ She also considers $L_p^w(G)$, $p>1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.