1
$\begingroup$

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?

$\endgroup$
5
  • 1
    $\begingroup$ None of these isomorphisms can hold for general $n$ for dimensional reasons: $\dim \mathfrak{u}(n) = n^2$ but $\dim \mathfrak{o}(n) = \frac{1}{2} n (n - 1)$. $\endgroup$ – Travis Willse Apr 1 '19 at 18:00
  • $\begingroup$ @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 $\mathfrak{gl}(n,\mathbb C)\simeq\mathfrak{u}(n)_{\mathbb C}$? $\endgroup$ – glS Apr 1 '19 at 18:06
  • $\begingroup$ I suppose that as a real vector space you can write this as $\mathfrak{u}(n) \cong \mathfrak{o}(n) \oplus \odot^2 \Bbb R^n$. $\endgroup$ – Travis Willse Apr 1 '19 at 18:23
  • $\begingroup$ And it's true that the vector space of symmetric real matrices is not a Lie algebra under the matrix commutator, as for symmetric $A, B$, $[A, B]^\top = (A B - B A)^\top = B^\top A^\top - A^\top B^\top = B A - A B = -[A, B]$. $\endgroup$ – Travis Willse Apr 1 '19 at 18:23
  • $\begingroup$ $\odot^k V$ 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 $\operatorname{Sym}^k V$. $\endgroup$ – Travis Willse Apr 1 '19 at 18:31
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.