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In some proofs of our lecture we used following relation $$\rho(g)u=u*\delta_g$$

The convolution was defined as $(u*v)(h)= \sum_{g\in\Gamma} u(g) v(g^{-1}h)$ for $u,v\in l^2(\Gamma)$ and $h\in\Gamma$ and the right regular representation as $(\rho(g)v)(h)= v(hg)$ for $v\in l^2(\Gamma)$ and $h\in\Gamma$

The problem is that my calculation yields another relation:

$$(u*\delta_g)(h)=\sum_{f\in\Gamma}u(f)\delta_g(f^{-1}h)$$

Every summand is zero, except the one that satisfies $f=hg^{-1}$ (since $g=f^{-1}h \Rightarrow f^{-1}=gh^{-1} \Rightarrow f=hg^{-1}$). Hence it follows

$$(u*\delta_g)(h)= u(hg^{-1})=(\rho(g^{-1})u)(h)$$

Have I made a mistake somewhere? Thanks for your help.

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The issue is in your formula for the right regular representation in terms of the convolution product: You state that $$\rho(g)u=u*\delta_g.$$ But the correct formula is $$\rho(g)u=u*\delta_{g^{-1}}.$$ With this, one obtains \begin{align*} (\rho(g)u)(h) &=(u*\delta_{g^{-1}})(h)\\ &=\sum_{f\in\Gamma}u(f)\delta_{g^{-1}}(f^{-1}h)\\ &=u(hg). \end{align*}

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  • $\begingroup$ Thanks for your help I also thought that possibly the definition is wrong in the lecture. But usually I am wrong and not the script. $\endgroup$ Aug 8 at 9:15
  • $\begingroup$ @Schrödinger'scat Ahh I'm so sorry, I got it backwards. Please see my correction. You can now check that $\rho(g_1g_2)=\rho(g_1)\rho(g_2)$, whereas the previous formula gave $\rho(g_1g_2)=\rho(g_2)\rho(g_1)$ for all $g_1,g_2\in\Gamma$. $\endgroup$
    – Aweygan
    Aug 8 at 12:24

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