## Local normal forms of covariantly constant selfadjoint (1,1)-tensors

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.

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 ...".

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).

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.

Does anybody know more?

-
 Slightly tangential: do you have a reference for the "linear algebra" counterpart for this question? That is given a nondegenerate quadratic form $g$ on a vector space $V$, and a $g$-self adjoint $L:V\to V$, what is the "normal form" for $L$? I only know the case for $g$ positive/negative definite or $g$ Lorentzian. – Willie Wong Apr 19 2011 at 11:54 I just checked the reference Matias Dahl has given and I did not know before: it gives a precise answer to your question (though it is possible to obtain this answer "by hand"). The rough answer is as follows: in a certain basis the matrix of $L$ has the standard Jordan normal form. The matrix of $g$ is also block-diagonal with the same dimensions of the blocks as the matrix of $L$; each block is (up to the sign) the anti-diagonal 1-matrix (i.e., the elements $a_{i , k-i+1}$ are 1 and other are 0). – Vladimir S Matveev Apr 19 2011 at 15:42 @willie: that was asked in MO a long time ago: mathoverflow.net/questions/1876 and was answered; although I didn't know the paper by Lancaster and Rodman. This is the sort of result that seems to have been "discovered" many times. – José Figueroa-O'Farrill Apr 20 2011 at 2:21 @Vladimir, Matias, Jose: thanks! Those are useful. The question from long time ago was really a long time ago! Before I even knew about MO! – Willie Wong Apr 20 2011 at 11:08

This does not answer the whole question, but addresses only the pointwise problem.

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. Lancaster, L. Rodman, Canonical Forms for Hermitian Matrix Pairs under strict Equivalence and Congruence, SIAM Review Vol 47, No 3, pp. 407--443.

-
 Thank you, Matias; I did not know this reference and therefore had to obtain the same local description by hands. – Vladimir S Matveev Apr 19 2011 at 15:36

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.

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.

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.

-
 Thank you, Robert. It may happen that I used wrong terminology in my question: instead of "normal forms" I should say "local description". I am indeed more interested in the answer of the form: "in a certain coordinate system, the pair $(g, L)$ are given by ...". So, from this point of view, the case $L=0$ is trivial, since in a certain coordinate system the metric is given by arbitrarily $g_{ij}(x)$ and $L$ is identically zero. P.S. It is indeed true that the most complicated cases are when $L$ has many Jordan blocks with the same eigenvalue and of approximately same dimensions. – Vladimir S Matveev Apr 19 2011 at 16:02 Dear Robert, I did today the calculation in a simpler case and observed the same phenomenon: the existence of one self-adjoint covariantly constant tensor implies the existence of additional properties that are differential and not algebraic consequences of this tensor. The simpler case is: the manifold is a flat cone over a 4D manifold of signature (2,2) and the g-selfadjoint (1,1)-tensor L has the property $L^2=0$ and has rank 2. Then, by calculations that required to go up to the second derivative of the curvature, one can show that the metric admits two lightlike parallel vectors. – Vladimir S Matveev May 4 2011 at 19:59 @Vladimir: Hmm, I'm not sure that I have understood the special case you mean. I just did the calculation for a pair $(g,L)$ on a $5$-manifold, where $L$ is $g$-symmetric, $L^2=0$, and $L$ has rank $2$. The general such structure depends on $3$ functions of $3$ variables and does not admit any parallel vector fields. The symmetry group of such a structure has dimension at most $9$, and this is achieved by some homogeneous structures that are not flat (and do not admit any parallel vector fields). I don't see how to get the general solution, but I can see how to get `most of the way' there. – Robert Bryant May 8 2011 at 20:25 @Robert: I added artificially the additional assumption that the metric is a cone metric, that is it has the form $dt^2 + t^2 g$, where $g$ is a metric on a 4D manifold. The assumption may sound artificial in this problem, but it is natural in the problem I was going to apply the possible answer on the question I have asked. Under this assumption (and also assuming $L^2=0$ and $rank(L)=2$, I calculated the components of $(g,L)$ in a certain coordinate system and to my surprise the image of $L$ is spanned over two parallel vector fields. There is no algebraic reasons for them though. – Vladimir S Matveev May 8 2011 at 21:48 @Robert: The case $L^2=0$ and $L^2= -id$ were solved in Kručkovič, G. I.; Solodovnikov, A. S. Constant symmetric tensors in Riemannian spaces. (Russian) Izv. Vysš. Učebn. Zaved. Matematika 1959 1959 no. 3 (10), 147–158. (I did not check the whole arguments, and actually did not look on the case $L^2= -id$ at all, but the part of the arguments I have checked was OK and the general method is also OK). – Vladimir S Matveev May 8 2011 at 21:54