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$\newcommand{\eS}{\mathscr{S}}$ $\DeclareMathOperator{\SO}{SO}$ $\newcommand{\eP}{\mathscr{P}}$ $\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\tr}{tr}$ Let $m>1$ be an integer and denote by $\eS_m$ the vector space of symmetric $m\times m$ real matrices. The group $\SO(m)$ acts by conjugation on $\eS_m$. The space of $\SO(m)$-invariant quadratic polynomials on $\eS_m$ is spanned by the two polynomials

$$ A \mapsto \tr A^2,\;\;A\mapsto (\tr A)^2. $$

Let $\SO(m-1)$ be the subgroup of $\SO(m)$ consisting of orthogonal transformations of $\bR^m$ that fix a unit vector $\eta$.

What is the space of $\SO(m-1)$-invariant on $\eS_m$. More precisely, can one explicitly write a basis of this space?

Clearly $\tr A^2$ and $(\tr A)^2$ are such polynomials, and so are

$$ A\mapsto (A^2\eta,\eta),\;\;A\mapsto (A\eta,\eta)^2, $$

where $(-,-)$ denotes the natural inner product on $\bR^m$. Are there any more $\SO(m-1)$ invariant quadratic polynomials? (I am inclined to believe that the above is the complete list.) Update After Robert Bryant's answer I lost my initial inclination.

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  • $\begingroup$ quite unrelated, but does that mean $q(A) = \alpha \operatorname{tr}(A^2) + \beta \operatorname{tr}(A)^2$ is the most general invariant quadratic form over the real symmetric matrices, and a linear combination of the five quadratic forms from the question and answer is the most general quadratic form that has ${\rm SO}(m - 1)$ invariance? $\endgroup$
    – user114668
    Commented May 18, 2020 at 8:11
  • $\begingroup$ @student Yes, that is correct $\endgroup$
    – Liviu Nicolaescu
    Commented May 18, 2020 at 9:01

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This is a trick question, right? It's not true when $m=2$ because, then $\mathrm{SO}(m{-}1)=\mathrm{SO}(1)$ is trivial, so that all polynomials on $2$-by-$2$ symmetric matrices are invariants, and the four quadratics you mention clearly don't span the the six-dimensional space of all quadratics.

Moreover, for $m>2$, you are missing $\mathrm{tr}(A)(A\eta,\eta)$. When $m>2$, these five do span the space of $\mathrm{SO}(m{-}1)$-invariant quadratic polynomials, as you can see by realizing that, when $m>2$, the space $\mathcal{S}_m$ of symmetric $m$-by-$m$ matrices splits under $\mathrm{SO}(m{-}1)$ into a direct sum $$ \mathcal{S}_m = \mathbb{R}\oplus\mathbb{R}\oplus\mathbb{R}^{m-1}\oplus \mathcal{S}^0_{m-1} $$ of $\mathrm{SO}(m{-}1)$-irreducible modules, where the only equivalences are between the first two (trivial) summands, and where $\mathcal{S}^0_{m-1}$ means the trace-zero symmetric $(m{-}1)$-by-$(m{-}1)$ matrices. (These two projections to $\mathbb{R}$ are given by the invariant linear forms $\mathrm{tr}(A)$ and $(A\eta,\eta)$.) [This splitting is valid for $m=2$, of course, but then the third summand is another copy of $\mathbb{R}$ and the fourth summand has dimension $0$.]

However, you switched from quadratic polynomials at the beginning to all polynomials. Was that intentional, or did you just want quadratics?

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  • $\begingroup$ @ Robert It was not a trick question. I wrote it early in the morning when my brain hadn't yet absorbed all the caffeine. I meant quadratic polynomials, and I now see that missed one. Thank you for the answer. $\endgroup$
    – Liviu Nicolaescu
    Commented Aug 15, 2012 at 14:32

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