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An "almost" true inequality for Hermitian matrices
@NarutakaOZAWA Thanks! This is a good point.
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An "almost" true inequality for Hermitian matrices
@NarutakaOZAWA This looks interesting, but it might be helpful if your notations ($e_k$, $f_i(k)$, overline, etc.) could be explained a little more.
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An "almost" true inequality for Hermitian matrices
@Malkoun That's a good point. In this case, A can be scaled to a projector. Whenever A is a projector, we have $A^p=A$ for any p, and this inequality always holds.
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What is the Lie superalgebra generated by permutations?
This is very awesome! It's quite surprising to me that the Lie subsuperalgebra generated by transpositions contains almost everything except for the centers.
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What is the Lie superalgebra generated by permutations?
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What is the Lie superalgebra generated by permutations?
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Softwares to determine semi-simple types of Lie algebras generated over $\mathbb{R}$ or $\mathbb{C}$ by a set of matrices
@მამუკაჯიბლაძე Thanks! The function
CanonicalBasis()
is helpful for me. It seems for this particular example it is easy to get the semisimple type over $\mathbb{R}$ by combining the results from over $\mathbb{Q}$ and over $CF(24)$. I guess GAP
would not be able to deal with matrices with transcendental elements, although I am not sure whether I will encounter such case.
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Softwares to determine semi-simple types of Lie algebras generated over $\mathbb{R}$ or $\mathbb{C}$ by a set of matrices
@მამუკაჯიბლაძე May I know how did you get this in
GAP
? Thanks!
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Prove: Lie algebra generated by two $n\times n$ shift matrices is $\mathfrak{so}(n,\mathbb{C})$ ($n$ odd) or $\mathfrak{sp}(n,\mathbb{C})$ ($n$ even)
This is very elegant. Thank you so much!
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Prove: Lie algebra generated by two $n\times n$ shift matrices is $\mathfrak{so}(n,\mathbb{C})$ ($n$ odd) or $\mathfrak{sp}(n,\mathbb{C})$ ($n$ even)
@abx Thanks for the suggestion! I've made the change in the text.
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