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It is a well-known result (proved for example also in this answer) that $\mathfrak{gl}(n,\mathbb C)\simeq \mathfrak{u}(n)_{\mathbb C}$, which can also be understood as another way to state that any complex $n\times n$ matrix can be uniquely decomposed as sum of a Hermitian and a skew-Hermitian matrix (equivalently, sum of two Hermitian matrices): $$A=\underbrace{\frac{A+A^\dagger}{2}}_{\text{Hermitian}}+i\underbrace{\frac{A-A^\dagger}{2i}}_{\text{Hermitian}}.$$ One can similarly show that any Hermitian matrix can be written uniquely as a sum of a symmetric and a skew-symmetric matrix. Indeed, if $H=H_R+iH_I$ is the decomposition of an arbitrary $H$ into real and imaginary part, then $H^\dagger=H$ if and only if $H_R^T=H_R$ and $H_I=-H_I^T$ (similarly, if $H$ is skew-Hermitian then $H_R$ is skew-symmetric and $H_I$ is symmetric).

This looks very close to the decomposition of a general matrix in terms of Hermitian and skew-Hermitian, but to decompose (skew-)Hermitian matrices in terms of real (skew-)symmetric matrices.

Can this isomorphism be stated on a similar footing as $\mathfrak{gl}(n,\mathbb C)\simeq \mathfrak{u}(n)_{\mathbb C}$? As noted in the comments, one difference in this case is that $\operatorname{dim}(\operatorname{Symm}(n))\neq \operatorname{dim}(\operatorname{skew-Symm}(n))$, and furthermore the set of symmetric real matrices does not have the structure of a Lie algebra (I think?). Still, is there a way around this? Does this bijection have interesting consequences?

glS
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    None of these isomorphisms can hold for general $n$ for dimensional reasons: $\dim \mathfrak{u}(n) = n^2$ but $\dim \mathfrak{o}(n) = \frac{1}{2} n (n - 1)$. – Travis Willse Apr 01 '19 at 18:00
  • @Travis mh, that is true, I missed the fact that while Hermitian and skew-Hermitian matrices have the same dimension, there are more symmetric than skew-symmetric matrices. Still, it is true that every element of $\mathfrak{u}(n)$ can be decomposed uniquely as sum of a symmetric and a skew-symmetric real matrix, right? Is there a way to state this similarly to how $\mathfrak{gl}(n,\mathbb C)\simeq\mathfrak{u}(n)_{\mathbb C}$? – glS Apr 01 '19 at 18:06
  • I suppose that as a real vector space you can write this as $\mathfrak{u}(n) \cong \mathfrak{o}(n) \oplus \odot^2 \Bbb R^n$. – Travis Willse Apr 01 '19 at 18:23
  • And it's true that the vector space of symmetric real matrices is not a Lie algebra under the matrix commutator, as for symmetric $A, B$, $[A, B]^\top = (A B - B A)^\top = B^\top A^\top - A^\top B^\top = B A - A B = -[A, B]$. – Travis Willse Apr 01 '19 at 18:23
  • $\odot^k V$ is a (somewhat) standard notation for symmetric tensor power of $V$. One might also write it (or $\bigodot^k V$) as $S^k V$ or $\operatorname{Sym}^k V$. – Travis Willse Apr 01 '19 at 18:31

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A standard physical consequence of this real symmetric / imaginary antisymmetric matrices split of hermitean matrices is the vanishing of structure constants in the Gell-Mann basis, illustrated here for $\mathfrak{su}(3)$: the 8 Lie algebra generators split into an imaginary antisymmetric set, $$ \lambda_2, \lambda_5, \lambda_7, $$ and a real symmetric set, $$ \lambda_1, \lambda_3, \lambda_4, \lambda_6, \lambda_8. $$

As an immediate consequence, the structure constants of the algebra $$ f^{ijk} = -\frac{1}{4} i \operatorname{tr}(\lambda_i [ \lambda_j, \lambda_k ]), $$ vanish unless their indices correspond to an odd number of indices from the set of odd generators, $\{ 2, 5, 7 \}$.

This generalizes readily to $\mathfrak{su}(n)$ and reminds you how sparse the set of structure constants is--not hard to work out the asymptotic density formula.

Cosmas Zachos
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