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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.

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