# Matt Pressland

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bio website people.bath.ac.uk/mdp33 location Bath, United Kingdom age 24 member for 2 years, 5 months seen 2 hours ago profile views 802

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

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 Apr4 comment What is Betti number of a group? One additional comment; I think the "number of generators" comment on Wikipedia is slightly wrong, but at the very least it means the minimal number of generators in a generating set, not the number of elements which are generators. So the "answer" for $\mathbb{Z}_6$ would be $1$, because it is generated by $1$ (or by $5$ - but one element is sufficient). Something like $\mathbb{Z}^2$ isn't generated by a single element; but it can be generated by two elements (e.g. $(1,0)$ and $(0,1)$), so the Betti number is $2$. Apr4 comment What is Betti number of a group? True - but the fact the OP says "Betti number" rather than "Betti numbers" suggests that this is not the point. I don't know if one of the cohomological Betti numbers will also agree with the rank. Apr4 revised What is Betti number of a group? added 69 characters in body Apr4 comment What is Betti number of a group? That page is about the Betti number for topological spaces; the $k$-th Betti number of the space is the rank of its $k$-th homology group, which deserves to be called the Betti number of the group! To calculate the homology group of a group, you need to put a topology on it; for finite abelian groups, the most obvious one is the discrete one, in which case the $0$-th Betti number is the number of elements, and all the others are $0$ - this probably isn't what you wanted! Apr4 answered What is Betti number of a group? Apr3 comment What does $s\in \{(u_0,u_1)\in \Bbb R \times\Bbb R^3\}$ mean? Saying subset is a little dangerous; elements of Cartesian products $A\times B$ are ordered pairs, whereas subsets (presumably of $A\cup B$ in this context) are unordered. This only really matters if you have $s\in A\times A$; the elements $(a_1,a_2)$ and $(a_2,a_1)$ of $A\times A$ are different, but the $2$-element subsets $\{a_1,a_2\}$ and $\{a_2,a_1\}$ of $A$ are the same. Apr2 revised For which values does the Gamma function yield an integer result? edited title Mar19 comment Why is $*$ defined only for homotopy classes, and not individual paths between points? Small typo: you mean "if $a=b$ and $c=d$ then $a*c=b*d$". I'm assuming you mean loops and not paths, and $*$ is the operation that says "do one loop and then the other", but maybe this is wrong...if you do mean this, then the operation is well-defined on paths, essentially automatically, because two paths are "equal" only when they are exactly the same. The more interesting statement is that it is well-defined on homotopy classes, because now some quite different looking paths are considered equal. Mar14 comment Finding remainder when $32^{32^{32}}$ is divided by $7$ A quick hint - you can use Fermat's little theorem to reduce the exponent modulo $6(=7-1)$ (because $a^{p-1}\equiv1\bmod{p}$ for any prime $p$). Mar11 comment Proving that $\dim(\mathrm{span}({I_n,A,A^2,…})) \leq n$ @clueless I decided to make a bold edit and replace the "diag" by "dim" (and $<$ by $\leq$ in the title), as it seems you agree that this is what it should be. It is better if anybody finds this page later if the question makes sense and agrees with the answer! However, if you don't agree with this edit, please roll it back (or if you don't know how, leave a comment and I'll do it myself.) Mar11 revised Proving that $\dim(\mathrm{span}({I_n,A,A^2,…})) \leq n$ deleted 9 characters in body; edited title Mar5 comment Does associativity justify $(f^{-1}gf)(f^{-1}hf) = f^{-1}gff^{-1}hf$? @foo1899 I guess my language possibly isn't clear. I mean that if you have no associativity, you don't know whether $fgh$ is supposed to mean $(fg)h$ or $f(gh)$, and it might matter. Hence you can't write $fgh$ (or any product of three elements) without establishing a convention about which is the correct reading. Mar5 comment Does associativity justify $(f^{-1}gf)(f^{-1}hf) = f^{-1}gff^{-1}hf$? @foo1899 It also isn't a priori defined: the fact that we can unambiguously write expressions consisting of the product of more than 2 elements without brackets is a consequence of associativity. (Also, SE should let you post an answer, so if you can't, this is a bug.) Mar5 comment Does associativity justify $(f^{-1}gf)(f^{-1}hf) = f^{-1}gff^{-1}hf$? (5) isn't really a well-defined statement because $fgh$ isn't a priori defined; $fgh$ may be defined to be equal to $(fg)h$, which is equal to $f(gh)$ by (2). This definition would be dangerous without associativity. The statement you want (where you drop all the brackets) follows by repeated application of associativity, but this may be a little fiddly to do in practice. (It would probably help to start by putting more brackets in, so that the elements all occur in bracketed pairs). Mar4 revised Do you know to find the sum of the series? Turned attached image into MathJax Mar4 revised Rigorous proof that any linear map between vector spaces can be represented by a matrix added 5 characters in body Mar4 comment A dense subset of a finite group The definition gives you the meaning of dense - the closure is $G$. The finiteness isn't relevant, this is purely topological. A subset $S$ of a topological space $X$ is dense if $\overline{S}=X$. Here you are taking the closure in the Zariski topology. Mar3 revised For the following function, determine whether it is a homomorphism. added 190 characters in body Mar3 comment For the following function, determine whether it is a homomorphism. OK - now you should check the definition of homomorphism. Take matrices $\left(\begin{smallmatrix}a&0\\b&a\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}c&0\\d&c\end{smallmatrix}\right)$, and multiply them together. If $\varphi$ takes the result to $\frac{b}{a}+\frac{d}{c}$, then you do have a homomorphism. I hope this last statement is clear, please ask if not! Mar3 comment For the following function, determine whether it is a homomorphism. You need to define $G$ and $R$ for us. If you already have some thoughts on the problem, it would be good to hear them as well.