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bio website people.bath.ac.uk/mdp33
location Bath, United Kingdom
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Postgraduate student at the University of Bath, UK, studying geometry and representation theory. Currently thinking about cluster algebras and related objects.


2d
comment Representation of $GL_2$ on $K^2$
If you follow what they say completely literally, you will need to invert all the matrices, which will make the expressions messier.
2d
comment Representation of $GL_2$ on $K^2$
In what I said, the induced action was a right action because the action on $K^2$ was a left action - in what you say two comments ago, the action $*$ on the domain is a right action. What you say one comment ago is correct. In terms of what to do about your problem, you will have to use the context of the problem to work out what they want. I wouldn't be surprised if they intended the action on $K[x,y]$ described in my answer, but this isn't consistent with the action induced by the standard left action on $K^2$.
Apr
11
comment Representation of $GL_2$ on $K^2$
Even more precisely; a left action of $G$ is "the same" as a right action of the opposite group $G^{\mathrm{op}}$, which has the same underlying set, but the operation is defined by $g\times_{\mathrm{op}}h=h\times g$, where $\times$ is the multiplication in $G$. But $G$ and $G^{\mathrm{op}}$ are isomorphic via $g\mapsto g^{-1}$, hence the relation between left and right actions in the comment above.
Apr
11
comment Representation of $GL_2$ on $K^2$
Not quite - you also need an inverse in the second expression, else you have the problem in my earlier comment. I'm not sure I would call them "exactly the same" either. A more precise statement is that each left action $\cdot$ induces a right action $*$ by defining $x*g=g^{-1}\cdot x$ (and vice versa) and the actions $g\cdot x(a,b)=g^{-1}(a,b)$ and $x\cdot g(a,b)=(a,b)g^{-1}$ are related in this way.
Apr
11
comment Representation of $GL_2$ on $K^2$
No - you have a left action on $K^2$, so it induces a right action on $K[x,y]$; if you had a right action it would induce a left action. Either way, they can't be on the same side - in what you just wrote, you would have $(x\cdot g)\cdot h(a,b)=x\cdot g((a,b)\cdot h)=x((a,b)\cdot hg)$, which isn't $x\cdot gh(a,b)$. You either have to have the actions on opposite sides, or introduce an inverse.
Apr
11
comment Representation of $GL_2$ on $K^2$
You extend the definition so that evaluation is a ring homomorphism - so $(2xy^2)(a,b)=2ab^2$. Moreover, $(2xy^2\cdot g)(a,b)$ is $2(g_{11}a+g_{12}b)(g_{21}a+g_{22}b)^2$, so $2xy^2\cdot g=2(g_{11}x+g_{12}y)(g_{21}x+g_{22}y)^2$.
Apr
11
answered Representation of $GL_2$ on $K^2$
Apr
10
answered Isomorphisms between groups
Apr
9
comment What are the consequences of presentation of an algebra by generators and relations?
A presentation of such an algebra as the path algebra of a quiver with admissible relations is a very useful one if you want to study its representation theory - this is a huge subject, but the book "Elements of the Representation Theory of Associative Algebras" by Assem, Simson and Skowroński covers lots of it.
Apr
7
revised Prove the order of an element divides the order of the group using cosets
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Apr
4
revised Prove the order of an element divides the order of the group using cosets
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Apr
4
answered Prove the order of an element divides the order of the group using cosets
Apr
4
comment Prove the order of an element divides the order of the group using cosets
That proof is valid - but I think you're just reproving Lagrange's theorem now. It may help to know what your statement of Lagrange's theorem is exactly, because there are different versions, some of which will be more immediately helpful than others.
Apr
4
comment Prove the order of an element divides the order of the group using cosets
If $H$ is the subgroup generated by $a$, what is the order (i.e. number of elements) of $H$?
Apr
4
comment What is Betti number of a group?
Said another way - in the textbook's language, the Betti number is not the number of generators, but the number of factors isomorphic to $\mathbb{Z}$ (and I agree that this is the better definition). As $\mathbb{Z}$ is not finite, no finite abelian group can have a factor isomorphic to $\mathbb{Z}$.
Apr
4
comment What is Betti number of a group?
Yes - this is why the number of generators comment on Wikipedia is wrong! The finite abelian group $\mathbb{Z}_5$ is generated by one object, but has Betti number zero (its free part is $0$). I think the confusion on Wikipedia is because they are talking about groups arising from topological spaces, and if you put enough restrictions on the space, you will only see groups isomorphic to $\mathbb{Z}^n$, so then the Betti number really is the number of generators - but this is not the case even for general topological spaces.
Apr
4
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 $2$.
Apr
4
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.
Apr
4
revised What is Betti number of a group?
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Apr
4
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!