Irreducible representation decomposition of tensor on manifold with metric - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T02:24:40Z http://mathoverflow.net/feeds/question/108940 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108940/irreducible-representation-decomposition-of-tensor-on-manifold-with-metric Irreducible representation decomposition of tensor on manifold with metric duetosymmetry 2012-10-05T17:44:50Z 2012-10-15T10:16:25Z <p>I'm aware that for some tensor product space, Schur-Weyl duality lets me decompose the space into irreducible representations by looking at irreps of the symmetric group. The simplest example is $V\otimes V\cong \mathrm{Sym}^2(V)\oplus \Lambda^2(V)$. In physicists' notation (warning, I am a physicist), we would write $T_{ab} = T_{(ab)} + T_{[ab]}$.</p> <p>However, when talking about tensors on a manifold $M$ with metric $g$, we can also take traces and so further reduce symmetric products into a trace and trace-free part, i.e. $T_{ab} = \frac{1}{d}g_{ab}T + [T_{ab}]^{\tiny{STF}} + T_{[ab]}$ where $T\equiv g^{ab}T_{ab}$, $d$ is the dimension of the manifold, and $[T_{ab}]^{\tiny{STF}} = T_{(ab)}-\frac{1}{d}g_{ab}T$. This seems to be relevant because because $g$ is an invariant symbol of $SO(p,q)$ where $(p,q)$ is the signature of $g$. There will also be an alternating tensor of highest rank which will be an invariant symbol, and this could also appear in the decomposition.</p> <p>Schur-Weyl doesn't seem to say anything about metrics, trace/trace-free decompositions, etc. What is the general decomposition here? Is there an algorithm to follow?</p> http://mathoverflow.net/questions/108940/irreducible-representation-decomposition-of-tensor-on-manifold-with-metric/108948#108948 Answer by Peter Michor for Irreducible representation decomposition of tensor on manifold with metric Peter Michor 2012-10-05T19:14:36Z 2012-10-06T17:37:48Z <p>Have a look at the beginning of section 33 (in particular, 33.2) of the book "Natural operations in differential geometry" (<a href="http://www.mat.univie.ac.at/~michor/kmsbookh.pdf" rel="nofollow">pdf</a>), for the Riemannian case only. It should work for the $SO(p,q)$ case also. There all $O(n)$-invariant tensors are described: The idea is to tensor with the metric or its inverse and then use the $GL(n)$ decomposition, i.e., involve traces and permutations.</p> <p>Edit: Link corrected.</p> http://mathoverflow.net/questions/108940/irreducible-representation-decomposition-of-tensor-on-manifold-with-metric/109703#109703 Answer by robot for Irreducible representation decomposition of tensor on manifold with metric robot 2012-10-15T10:16:25Z 2012-10-15T10:16:25Z <p>There is a "Schur-Weyl theory" for representation of $O(n)$ and $Sp(n)$. The group algebra of the symmetric group is replaced by the <a href="http://en.wikipedia.org/wiki/Brauer_algebra" rel="nofollow">Brauer algebra</a>. Basically, you first decompose your tensor product with respect to "the number of traces the vectors contain" and then for each such part you can use the classical $GL$ decomposition. </p> <p>The relevant representation theory over complex numbers is treated in a book by Wallach and Goodman <a href="http://www.math.rutgers.edu/~goodman/repbook.html" rel="nofollow">Symmetry, Representations, and Invariants</a>. In particular look at the appendix F on the linked web page. Alternatively, you can look into the older version of this book which was published under the name Representations and Invariants of the Classical Groups. </p> <p>You should be careful when dealing with representation of noncompact real groups. </p>