Convolution on symmetric group Sn - MathOverflow most recent 30 from http://mathoverflow.net2013-06-18T22:18:37Zhttp://mathoverflow.net/feeds/question/57276http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/57276/convolution-on-symmetric-group-snConvolution on symmetric group SnDeepti Pachauri2011-03-03T18:43:19Z2011-07-31T08:53:59Z
<p>I have question regarding convolution of functions (say
g and h) defined on Sn. In Fourier space this is equivalent to IFT(G.H),
where G = FT(g) and H = FT(h).</p>
<p>Fast Fourier transforms (Clausen's FFT) proceeds by recursively breaking
down Fourier transformation over Sn into smaller transforms over S_(n-1),
S_(n-2)... and computing each S_(k)-transform from the k independent
S_(k-1) transforms.</p>
<p>Now the question I have is - How does the convolution of two functions (g
& h, each defined on Sn) translate to S_(n-1)? In other words, is their
any defining expression involving G' and H' to provide the n-1 independent
S_(n-1) transforms to get final the convolution.</p>
<p>G': descendant Fourier transform of G on S_(n-1)
H': descendant Fourier transform of H on S_(n-1)
FT: Fourier transform
IFT: Inverse Fourier transform</p>
<p>I would appreciate if anyone can direct me to some papers/books which talk
about these concepts.</p>
<p>DP</p>
http://mathoverflow.net/questions/57276/convolution-on-symmetric-group-sn/57293#57293Answer by Klim Efremenko for Convolution on symmetric group SnKlim Efremenko2011-03-03T22:29:08Z2011-03-03T22:44:36Z<p>It is not an answer to your question, but I hope it will help:</p>
<p>In general arithmetic complexity of convolution in non-anelian groups "equivalent" to the complexity of matrix multiplication.
Here is the reason why:</p>
<p>The way of doing Fourier Transform in abelian group $A$ can be described in the is the following way: Let $f,g \in F[A]$
We know that $F[A]$ is isomorphic to the space $F^A$ with pointwise multiplication.
Let $T$(which is acctually Fourier Transform) be this isomorphism. If we want calculate $f*g$ then calculate $T^{-1}(T(f)\cdot T(g))$. In case of non abelian group like $S_n$ It holds that
$F[G]$ is isomorphic to the direct sum of matrix algebras that is $F[G]\simeq\oplus M_{n_i}$. Thus using the same formula you can calculate convolution in $S_n$, but now you will need to multiply matrixes.</p>
http://mathoverflow.net/questions/57276/convolution-on-symmetric-group-sn/71717#71717Answer by openID for Convolution on symmetric group SnopenID2011-07-31T08:53:59Z2011-07-31T08:53:59Z<p><a href="http://www.cbse.ucsc.edu/sites/default/files/convolutions.pdf" rel="nofollow">http://www.cbse.ucsc.edu/sites/default/files/convolutions.pdf</a>
http://www.mpi-hd.mpg.de/astrophysik/HEA/internal/Numerical_Recipes/f13-1.pdf</p>
<p>might help you </p>