Invariants of symmetric matrices - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T19:07:47Z http://mathoverflow.net/feeds/question/104751 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104751/invariants-of-symmetric-matrices Invariants of symmetric matrices Liviu Nicolaescu 2012-08-15T08:59:10Z 2012-08-15T16:25:01Z <p>$\newcommand{\eS}{\mathscr{S}}$ $\DeclareMathOperator{\SO}{SO}$ $\newcommand{\eP}{\mathscr{P}}$ $\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\tr}{tr}$ 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</p> <p>$$A \mapsto \tr A^2,\;\;A\mapsto (\tr A)^2.$$</p> <p>Let $\SO(m-1)$ be the subgroup of $\SO(m)$ consisting of orthogonal transformations of $\bR^m$ that fix a unit vector $\eta$.</p> <p>What is the space of $\SO(m-1)$-invariant on $\eS_m$. More precisely, can one explicitly write a basis of this space?</p> <p>Clearly $\tr A^2$ and $(\tr A)^2$ are such polynomials, and so are</p> <p>$$A\mapsto (A^2\eta,\eta),\;\;A\mapsto (A\eta,\eta)^2,$$</p> <p>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.) <strong>Update</strong> After Robert Bryant's answer I lost my initial inclination.</p> http://mathoverflow.net/questions/104751/invariants-of-symmetric-matrices/104762#104762 Answer by Robert Bryant for Invariants of symmetric matrices Robert Bryant 2012-08-15T12:49:11Z 2012-08-15T12:49:11Z <p>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. </p> <p>Moreover, for $m>2$, you are missing $\tr(A)(A\eta,\eta)$. 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 <code>$\mathcal{S}_m$</code> of symmetric $m$-by-$m$ matrices splits under $\mathrm{SO}(m{-}1)$ into a direct sum <code>$$\mathcal{S}_m = \mathbb{R}\oplus\mathbb{R}\oplus\mathbb{R}^{m-1}\oplus \mathcal{S}^0_{m-1}$$</code> of <code>$\mathrm{SO}(m{-}1)$</code>-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 $\tr(A)$ 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 $0$.]</p> <p>However, you switched from quadratic polynomials at the beginning to all polynomials. Was that intentional, or did you just want quadratics?</p>