Local normal forms of covariantly constant selfadjoint (1,1)-tensors - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T01:12:15Z http://mathoverflow.net/feeds/question/62244 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62244/local-normal-forms-of-covariantly-constant-selfadjoint-1-1-tensors Local normal forms of covariantly constant selfadjoint (1,1)-tensors Vladimir S Matveev 2011-04-19T10:31:44Z 2011-04-20T20:25:58Z <p>Consider the pair $(g, L)$, where $g$ is a pseudo-Riemannian (i.e., nondegenerate and of arbitrary signature) metric and $L$ is an (1,1) $g$-selfadjoint tensor field. </p> <p>Does there exist a local description of such pairs; for example in the form "in a certain coordinate system $g$ and $L$ are given by formulas ...". </p> <p>For me, the state of art is as follows: for Riemannian metrics, the answer was known to classics: The existence of such $L$ implies the local decomposition of $g$ into a direct product: $g= g_0+ ... + g_k$, where $g_0$ is flat (we can think therefore that $g_0= dx_1^2 +...+dx_m^2$ in a certain coordinate system) and each $g_i$ has irreducible holonomy group. For such metrics $g= g_0+ ... + g_k$, the tensor $L$ also can be decomposed into the product $L= L_0+...+L_k$; moreover, $L_0$ is given by arbitrary symmetric matrix with constant entries and other $L_i$ are proportional to identity with constant coefficients (the coefficients depend on the component). </p> <p>For the pseudo-Riemannian metrics one can do the same splitting if $L$ has different eigenvalues; so the interesting case is when $L$ has one real eigenvalue or two conjugated complex eigenvalues. And this case looks quite open for me; only the special case when $L^2= 0$ or $L^2= -1$ are known.</p> <p>Does anybody know more? </p> http://mathoverflow.net/questions/62244/local-normal-forms-of-covariantly-constant-selfadjoint-1-1-tensors/62280#62280 Answer by Matias Dahl for Local normal forms of covariantly constant selfadjoint (1,1)-tensors Matias Dahl 2011-04-19T15:08:38Z 2011-04-19T15:08:38Z <p>This does not answer the whole question, but addresses only the pointwise problem.</p> <p>Let $x^i$ be local coordinates around point a point $p$ and let $B=(g_{ij})_{ij}$ and $A=(L^j_i)_{ij}$ be matrices that represent $g$ and $L$ at point $p$. Then $$\det B \neq 0, \quad B=B^t, \quad A^t B = B A.$$ Such a pair $A,B$ has a canonical form theorem similar to the Jordan normal form theorem. That is, there is a real invertible matrix $S$ such that matrices $S^{-1} A S$ and $S^T B S$ have normal forms. See Theorem 12.2 in:</p> <p>P. Lancaster, L. Rodman, Canonical Forms for Hermitian Matrix Pairs under strict Equivalence and Congruence, SIAM Review Vol 47, No 3, pp. 407--443.</p> http://mathoverflow.net/questions/62244/local-normal-forms-of-covariantly-constant-selfadjoint-1-1-tensors/62286#62286 Answer by Robert Bryant for Local normal forms of covariantly constant selfadjoint (1,1)-tensors Robert Bryant 2011-04-19T15:34:47Z 2011-04-20T20:25:58Z <p>Well, I don't have the complete answer, but then I don't think that a 'complete' answer is going to be simple. For example, you haven't ruled out the case $L=0$, which amounts to giving a normal form' for all pseudo-Riemannian metrics. What is true, of course, is that the algebraic type of $(g,L)$ is the same at all points (assuming that $M$ is connected) and, as Willie Wong points out, you will at least have to classify those algebraic types as a starting point.</p> <p>Of course, you can get complete answers in some algebraic types. For example, when $L$ has a single real eigenvalue $r$ (which must be constant), you can assume that this eigenvalue is $0$ by replacing $L$ by $L-rI$, and the least degenerate' case would presumably then be when $L^n=0$ but $L^{n-1}\not=0$. (I'm assuming that the dimension of the manifold is $n$.) Then it is not difficult to show that there is a (local) frame field $e_1,\ldots,e_n$ such that $e_i = L^{i-1}e_1$ for $i = 2,\ldots,n$ and, moreover, such that $g(e_i,e_{n+1-i}) = 1$ with all other $g(e_i,e_j) = 0$. (You may have to replace $g$ by $-g$ in order to arrange this, but that's a trivial matter.) Moreover, this frame field is unique up to replacing $e_i$ by $-e_i$ for all $i$. Since $L$ is $g$-parallel, it follows that this frame field is also $g$-parallel. In particular, the connection is flat and one can (locally) choose coordinates $x^i$ such that $dx^i(e_j) = \delta^i_j$. In these coordinates, $g$ and $L$ have constant coefficients.</p> <p>Added after Vladimir's comment: Yes, it appears that the case of more than one Jordan block is more interesting. I did a back-of-the-envelope calculation for the case of a $5$-dimensional manifold $M$ with a pair $(g,L)$ where $L$ is nilpotent with $2$ Jordan blocks, one of size $3$ and one of size $2$. The result is that there are $4$ algebraic types possible ($2$ if you allow the replacement of $g$ by $-g$), and each of them exists as a $1$-parameter family of inequivalent types. One member in each family is 'flat', i.e., $g$ and $L$ have constant coefficients in the appropriate coordinate system, but the others aren't flat, even though they are homogeneous (with a $7$-parameter family of $L$-preserving isometries). Hmm.</p>