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Feb
2
answered Why is the Young symmetrizer non-zero?
Feb
1
awarded  Popular Question
Jan
9
comment Has this equation appeared before?
I have accepted your answer, although I haven't verified it in Mathematica yet; by hand I can only verify the special case $a=b=c$.
Jan
9
accepted Has this equation appeared before?
Jan
8
comment Has this equation appeared before?
Reference for the above fact? Is it mathematica again :)? and what is the Galois group for our case (which should be solvable)?
Jan
8
comment Has this equation appeared before?
Wow!, thanks. How did you factor the degree 8 polynomial? Did you use brute force or a computer program, or did you use any clever technique?
Jan
5
comment Character related to a maximal subgroup
Yes, so $g\in H$ implies $\xi (g)\ne 0$, that is, elements of $\ker (\xi)$ can not be in $H$.
Jan
4
comment Has this equation appeared before?
@Tom-Tom: Nope, plane (2D) geometry, though I am more interested in the above algebraic equation.
Jan
4
comment Has this equation appeared before?
It appeared while I was working with a geometric problem.
Jan
4
asked Has this equation appeared before?
Jan
4
comment Character related to a maximal subgroup
If some element $g$ of $G$ fix every coset, then doesn't it act as identity permutation on the coset space, i.e. $\xi (g)=\mathrm{trace}(id )=|G/H|\ne 0$?
Jan
4
comment Character related to a maximal subgroup
That's my question, how is $\ker\xi\subseteq H$?
Jan
4
comment Character related to a maximal subgroup
May be I am missing something, but, isn't $\xi$ is same as $\mathbb{C}[G/H]$ as $G$-module ($G$ acts by permuting the cosets) and so $\ker \xi\subseteq G-H$ (since every element of $G$ other than $H$ moves every coset).
Jan
1
comment A matrix defines a self adjoint operator if and only if it is symmetric
@Omnomnomnom Both follows from here, the other implication should follow from the fact that $Mx=0$ for all $x$ implies $M=0$, I left this as exercise.
Jan
1
revised A matrix defines a self adjoint operator if and only if it is symmetric
added 26 characters in body
Jan
1
answered A matrix defines a self adjoint operator if and only if it is symmetric
Oct
19
awarded  Taxonomist
Oct
9
comment Dimension of the space of cubic polynomials over $\mathbb{P}^5$ which vanish on the Veronese surface.
I am not familiar with this stuff, can you tell me the reference where I can find the algorithm that has been used here.
Oct
8
asked Dimension of the space of cubic polynomials over $\mathbb{P}^5$ which vanish on the Veronese surface.
Oct
8
awarded  Nice Answer