Timeline for Local normal forms of covariantly constant selfadjoint (1,1)-tensors
Current License: CC BY-SA 3.0
7 events
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| May 8, 2011 at 21:54 | comment | added | Vladimir S Matveev | @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). | |
| May 8, 2011 at 21:48 | comment | added | Vladimir S Matveev | @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. | |
| May 8, 2011 at 20:25 | comment | added | Robert Bryant | @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. | |
| May 4, 2011 at 19:59 | comment | added | Vladimir S Matveev | 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. | |
| Apr 20, 2011 at 20:25 | history | edited | Robert Bryant | CC BY-SA 3.0 |
added a example with 2 Jordan blocks
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| Apr 19, 2011 at 16:02 | comment | added | Vladimir S Matveev | 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. | |
| Apr 19, 2011 at 15:34 | history | answered | Robert Bryant | CC BY-SA 3.0 |