# Relation between right regular representation and convolution of group algebras

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.

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*}
• @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 $g_1,g_2\in\Gamma$. Aug 8 at 12:24