# Tag Info

3

The natural surjection $A\to A/\text{rad}A$ induces a surjection $A^\times\to (A/\text{rad}A)^\times$, and so $(A/\text{rad}A)^\times$ is path connected. $A/\text{rad}A$ is a finite-dimensional commutative semisimple $\mathbb{R}$-algebra, and so is a finite product of copies of $\mathbb{R}$ and $\mathbb{C}$, and since its group of units is path connected ...

3

I believe you have the wrong copy of $\mathbb{C}$ inside $\mathbb{C}[G]$. Instead, you want to take $\mathbb{C}\cong\{c\cdot\delta_e\mid c\in\mathbb{C}\}$, where $\delta_e$ is the function $$\delta_e(g)=\begin{cases}1&\mbox{if }g=e\\0&\mbox{otherwise.}\end{cases}$$ In fact, there is an isomorphism of algebras ...

3

To explain what I said on MO (and also explaining what goes on in knsam's answer): The way to think about this is that since the group permutes the given basis vectors, it fixes the sum of all the given basis vectors. This gives a $1$-dimensional invariant submodule.

2

Hint. Arguably, the simplest invariant subspace would be one of dimension $1$. What would such a thing be? Do you see any such subspace in this case?

1

Every unitary matrix is diagonalizable, but $\rho(x)$ is not diagonalizable unless $x=0$ (its only eigenvector (up to scaling) is $(1,0)$).

1

The integral sign comes by fiat. The fundamental theorem of calculus tells us that the integral of a continuous function is (continuously) differentiable, introducing an integration gives us a more regular function to work with. While we only know that $\pi$ is continuous, we know that the integral of $\pi$ is differentiable, and therefore we introduce the ...

Only top voted, non community-wiki answers of a minimum length are eligible