3

Hint: The space of $4\times 4$ symmetric traceless matrices is of dimension $9$.


2

In fact, knowing the unitary matrix $e^{iA}$, we can explicitly calculate the associated matrices $A$. Clearly, we may put $t=1$ and we may assume that $spectrum(A)=(\lambda_j)_j$ and (up to an orthonormal change of basis) $e^{iA}=diag(e^{i\mu_1}I_{k_1},\cdots,e^{i\mu_r}I_{k_r})$, where the $e^{i\mu_j}$ are the distinct values of the $e^{i\lambda_j}$. $|e^{...


2

Presumably $t$ is a nonzero real number. Since matrix exponentials of non-trivial Jordan blocks are not diagonalisable over $\mathbb C$, if $e^{itA}$ is unitary, $A$ must be diagonalisable and every eigenvalue $\lambda$ of $A$ satisfies $|e^{it\lambda}|=1$. Hence each $\lambda$ is real, i.e. $A$ is a diagonalisable matrix with real eigenvalues, and $|\det(e^{...


2

[Begin EDIT] The following answer is correct under the additional assumption that the matrix $A$ is "nice" enough. As such, it describes a set of matrices with this property, but not all of them. I leave the answer unedited below, and add a discussion of a counterexample below. [End EDIT] I am assuming (from context clues) that $t\in\mathbb R$. Now $e^{...


2

Here are two different approches. First, one can define an action of $H'$ on $\tilde{G}/\overline{H}$ as follows. Given $g\in \tilde{G}$ and $h\in H'$, set $h\ast g\overline{H} = (gh^{-1})\overline{H}$. In general, this kind of "right multiplication" isn't well defined. But I claim that is is in this case, owing to the fact that $\overline{H}$ is a ...


1

Very short answer: You have to be very precise about which base field, $\mathbb R$ or $\mathbb C$, you are considering in each case. Over $\mathbb C$, there is the Lie group $SL_n(\mathbb C)$ and its Lie algebra $\mathfrak{sl}_n(\mathbb C)$, and every Cartan subalgebra of this will have roots which form a system of type $A_{n-1}$. There is extensive ...


1

The group $SL(n,\mathbb C)$ is a complex Lie group whose Lie algera is $\mathfrak{sl}(n,\mathbb C)$. The group $SU(n)$ is a compact real Lie group whose Lie algebra is $\mathfrak{su}(n)$, the Lie algebra of all skew-symmetric $n\times n$ complex matrices with null trace. It turns out that its complexification (that is, $\mathfrak{su}(n)\bigotimes\mathbb C$) ...


1

Almost every treatment of representations of real semisimple Lie algebras explicitly treats the basic case $\mathfrak{su}_2$; the standard result is that for each positive integer $n$, there is up to equivalence exactly one irreducible $n$-dimensional $\mathfrak{su}_2$-representation, call it $\rho_n$. Now if you already know $\mathfrak{so}_4 \simeq \...


1

For finite fields $F$, you will find a very detailed and comprehensive treatment in the book by Kleidman and Liebeck on the maximal subgroups of the finite classical groups.


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