# Matt Pressland

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bio website people.bath.ac.uk/mdp33 location Bath, United Kingdom age 25 member for 2 years, 6 months seen Aug 11 at 15:14 profile views 811

Postgraduate student at the University of Bath, UK, studying geometry and representation theory. Currently thinking about cluster algebras and related objects.

# 1,916 Actions

 Jun28 reviewed Edit suggested edit on Matrix similar to its inverse Jun28 revised Matrix similar to its inverse tex markup Jun28 comment Are $\mathbb{C} \otimes _\mathbb{R} \mathbb{C}$ and $\mathbb{C} \otimes _\mathbb{C} \mathbb{C}$ isomorphic as $\mathbb{R}$-vector spaces? Moreover, as hinted at by your first bullet, $C\otimes_{\mathbb{C}}\mathbb{C}\cong\mathbb{C}$, as $z_1\otimes z_2=z_1z_2\otimes 1$ for any $z_1,z_2\in\mathbb{C}$. Jun28 comment Does a smooth map from two disjoint spheres to one sphere exist? Any constant map between any two smooth manifolds is smooth, so I assume you want your maps to have more properties than just smoothness. There exist smooth injective maps from the disjoint union of two spheres to a single sphere, but not diffeomorphisms. Jun28 answered What exactly is a vector bundle isomorphism Jun27 revised degree of the Hilbert polynomial of a quotient added 2 characters in body Jun27 comment Is the number of irreducibles in any number field infinite? @pritam OK, that sounds good. (I suppose despite not knowing anything about this type of ring, I should at least have guessed that they are never fields). Jun27 comment Is the number of irreducibles in any number field infinite? One note that might be helpful - if there are finitely many units, then the argument goes through by your last sentence. You should probably also assume there is at least one irreducible as well. I'm thinking in terms of general rings, so maybe some of this is covered by the assumption that it's the ring of integers of an algebraic number field (I'm not a number theorist, so I don't really have a good feel for those). Jun19 comment How useful are geometric aspects when studying finite groups? Tangentially, there is lots of amazing (non-GGT) geometry coming from finite subgroups of $\operatorname{SL}(2,\mathbb{C})$ (and even $\operatorname{SL}(3,\mathbb{C})$); look up the McKay correspondence if you want to know more. Jun18 revised Matrix which commutes with permutation matrix added 313 characters in body Jun18 comment Matrix which commutes with permutation matrix But that doesn't matter - what you want to prove is the third equation in your answer, and you can obtain that from the result for transpositions by commuting $A$ past one of the transpositions at a time, i.e. $P[2,1,3]P[3,2,1]A=P[2,1,3]AP[3,2,1]=AP[2,1,3]P[3,2,1]$. The problem from the failure of commutativity of the permutations is that you can't just apply the same permutation to the rows of $A$ and to the columns and compare the answers, because that doesn't correspond to multiplication by the same permutation matrix on each side. Jun18 comment Matrix which commutes with permutation matrix @ZettaSuro You convinced me for a bit, but in fact it does - when you actually write it out you see that $P[3,1,2]$ acts on the right by applying the permutation $[3,1,2]$ to the columns, but on the left by applying the permutation $[2,3,1]$ to the rows. This is probably the source of the OP's confusion (and it confused me too) - only transpositions and the identity act by the "same" permutation on both sides. I now think my argument works (although Quimey's is better). Jun18 revised Matrix which commutes with permutation matrix deleted 48 characters in body Jun18 comment Matrix which commutes with permutation matrix @user17574 Still not quite right - I think the problem is that you can't use the word "general" quite like that - there's no such thing as "a general permutation matrix", but there is such a thing as "all permutation matrices". I'll edit it and you can change it again if you're unhappy with my phrasing. Jun18 answered Matrix which commutes with permutation matrix Jun18 comment Matrix which commutes with permutation matrix @user17574 That's not what you've literally said in the first sentence of your question (although I read what I expected to see, so didn't notice until the comment). Jun14 reviewed Edit suggested edit on Given $f(x+y)=f(x)f(y), f'(0)=11,f(3)=3$, what is $f'(3)$? Jun14 revised Given $f(x+y)=f(x)f(y), f'(0)=11,f(3)=3$, what is $f'(3)$? made the display proper Jun12 comment In diagonalization, can the eigenvector matrix be any scalar multiple? @Ian It's not too strange - see MyUserIsThis's answer. If $v$ is an eigenvector, so is any scalar multiple $cv$ (because if $Av=\lambda v$ then $A(cv)=cAv=c\lambda v=\lambda (cv)$. So any multiplying the eigenvectors by any constant (it doesn't have to be the same one for each eigenvector either) gives you another basis with the same properties as far as the action of $A$ goes. Jun12 comment Confusion regarding direct sum decomposition of representations from Serre's book If the space you're direct summing copies of is non-zero, then it doesn't make sense to write an internal direct sum anyway, so it must be an external direct sum in some sense. Note that in this example, we can say $V_i\cong W_i\oplus\cdots\oplus W_i$ where we mean an external direct sum of $W_i$s, and then the choice of isomorphism allows us to translate that into a decomposition of $V_i$, each component of which is isomorphic to $W_i$. (I guess all I'm really saying is that internal and external direct sums are isomorphic when both defined).