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bio website people.bath.ac.uk/mdp33
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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.


Apr
23
answered Matrix representations of Transformation with change of basis (Fraleigh Beauregard)
Apr
23
comment Matrix representations of Transformation with change of basis (Fraleigh Beauregard)
OK - that makes more sense. (In my previous comment, $R_{B,B''}$ should just say $R_{B''}$.)
Apr
23
comment Matrix representations of Transformation with change of basis (Fraleigh Beauregard)
Oh, that's probably not a useful way of thinking about it (although it is true). The procedure is to write $T(v_j)=\sum_{i=1}^n\lambda_{ij}v_i$ for some constants $\lambda_j$, for each $v_j$ in the basis $B$, and then the $j$-th column of $R_B$ consists of the $\lambda_{ij}$s. So each column should tell you the action of the transformation on a single basis vector. I think there might be a typo in the second example; $B$ and $B'$ are the same, and $R_B$ isn't correct. This matrix is actually $R_{B,B''}$, where $B''=(x^2,x,1)$.
Apr
23
comment Matrix representations of Transformation with change of basis (Fraleigh Beauregard)
The $R_B$ and $R_{B'}$ could be written more fully as $R_{B,B}$ and $R_{B',B'}$ to match the notation from before - this time the transformation is from a vector space to itself, so it makes more sense to only choose one basis. (I don't know if this is what you meant when you said you weren't sure what $R_B$ and $R_{B'}$ represent, but maybe it's helpful to say anyway!).
Apr
23
comment Matrix representations of Transformation with change of basis (Fraleigh Beauregard)
I don't see what you mean - it looks vertical in the second example; the third entry in the first column of the augmented part is $0$, which is the third component of $T([1,1,0])$, not the first component of $T([0,1,1])$. The $R_{B,B'}$ are matrices of a linear transformation with respect to the bases $B$ and $B'$, I'm not aware of a shorter name than that.
Apr
22
comment Permutations expressed as product of transpositions
Ah, you made a different mistake that gives the same result as doing the transpositions backwards - on the third arrow you should be swapping the numbers $1$ and $4$ over, wherever they appear, not the first position with the fourth. (I was drawing my way of doing this calculation; put $1$ in on the right and see where it goes - it first gets mapped to $4$ (by $(1,4)$), then $4$ is mapped to $6$ by $(4,6)$, and finally $6$ is mapped to $2$ by $(2,6)$).
Apr
22
comment Permutations expressed as product of transpositions
I'm not sure I understand your first question - but for the second one, you're just reading the transpositions in the wrong order. Your convention for $\sigma_1\circ\sigma_2$ is that $\sigma_2$ is applied first, so with the transpositions you need to start from $(3,6)$ and read from right to left. So $1\mapsto 4\mapsto 6\mapsto 2$ etc.
Apr
22
answered What does $f: 2^{\mathcal{S}}\rightarrow\,\mathbb{R}$ mean?
Apr
17
revised Why we can consider both modules as modules over $R_{(p)}$? (Bruns and Herzog, Theorem 1.5.9)
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Apr
14
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.
Apr
14
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
deleted 7 characters in body