# Can $\mathfrak{u}(n)$ be decomposed as direct sum of the sets of symmetric and skew-symmetric real matrices?

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?

• 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 1 '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 1 '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 1 '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 1 '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 1 '19 at 18:31

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.