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comment permutation representation of Symmetric group
If you know about Specht module, then it is easy exercise to see that this representation (called natural representation) can be decomposed into $S^{(n)}$ and $S^{(n-1,1)}$. Discretely $S^{(n)}=\mathbb{C}-span\{v_1+\cdots+v_n\}=$ trivial representation, where $v_1,\ldots,v_n$ is the standard basis of the natural rep. $S^{(n-1,1)}$ is the kernel of the projection of the natural representation project onto the trivial representation. The natural rep is the induction of trivial representation as rep of trivial group to $S_n$, so it is a permutation representation by definition.
Nov
3
comment How to compute $Ext_A^{1}(S_1, S_2)$ and $Ext_A^{1}(S_2, S_1)$?
By the way, in general, whether to count $i\to j$ or $j\to i$ depends on how you define Ext-quiver, or which way you work (i.e. left modules or right modules, and direction of multiplying arrows). In your example, we see from how you write down $P(1)$ to determine which direction is the correct one.
Nov
3
answered Compatibility of homomorphisms and quotient maps of abelian groups
Nov
2
comment Compatibility of homomorphisms and quotient maps of abelian groups
How (1) is used is the same (dual) thing of my previous comment: you need $h$ to be well-defined, i.e. any difference in $a$ by an element in $A'$ will give you the same element in the image. Here you will need to use the commutative square of the left.
Nov
2
comment Compatibility of homomorphisms and quotient maps of abelian groups
$g(a'):=f(a')$ is not quite the correct way as the element is in $C$, not in $C'$ by trivial reason. You should think, I have the diagram with the lower left corner taken out, and I have element in $A'$, how to get to some element in $C'$? Well, if you can push the element all the way to lower right corner ($C/C'$) and show that the result element is $[0]=0+C'\in C/C'$, that means you have something lying in $C'$.
Nov
2
comment How to compute $Ext_A^{1}(S_1, S_2)$ and $Ext_A^{1}(S_2, S_1)$?
I will just clarify Mariano and Julian's comment for the OP. The mistake in the calculation is that the ``projective resolution" you quoted works only for module of projective dimension 1. If you actually calculate the kernel of $P_1\to S_1$, you will see it is $S_2$, so the resulting ses is $0\to S_2\to P_1\to S_1\to 0$. In particular, the non-zero element of $Ext_A^1(S_1,S_2)$ corresponds to this non-split extension. If you only want the dimension of the Ext-groups between simples, then all you need to do is to count the number of arrows from one vertex to another -no calculation needed.
Oct
27
awarded  Revival
Oct
24
comment Semisimple modules and the radical
What is your condition on $A$? If $A$ is finite-dimensional algebra, then $\text{rad}V=\text{rad}AV$, and since $A/\text{rad}A$ semisimple, so is $V/\text{rad}V$.
Oct
18
comment Representation Theory Symmetric Group Book?
There are also Segal's book - also the classical approach to Specht modules like James-Kerber. Okuonkov-Vershik have a book on their approach to symmetric group, which differ slightly from the James-Kerber school approach, but enlightened many modern approach to KLR algebras, also used by Mathas and Kleshchev.
Oct
18
comment Representation Theory Symmetric Group Book?
James' Springer lecture notes and James-Kerber book is the usual recommended reference; but one may not like it because they discuss methods from decades ago. Mathas book on Iwahori-Hecke algebra is also good but computational heavy, in that sense, it may be "too explicit" for people. I personally like Kleshchev lecture notes, there are several versions depending on which direction (purely representation and homological within the realm of symmetric group, or more into generalisation to Hecke and KLR algebras).
Oct
9
comment Matrices of subrepresentations and quotient representations.
@Artos: Yes, but there is no guarantee that you can pick a subset of a chosen basis of $V$ to give a subrepresentation, even if a proper subrepresentation exists. An example, the regular representation of the cyclic group of order 2, over $\mathbb{C}$ (not Lie algebra, but group algebra, still serve the purpose) has a basis $1,\sigma$ (corresponding to elements of the group. But can't find a subset of this basis to form a subrepresentation, but there are two 1-dimensional subrepresentations: $\mathbb{C}(1+\sigma)/2$ and $\mathbb{C}(1-\sigma)/2$.
Oct
9
comment Matrices of subrepresentations and quotient representations.
Yes and no. Yes, if you can truly find a basis of $V$ so that for all $x\in\mathfrak{g}$, $\rho_x$ is a 5x5 matrix with $(\rho_x)_{ij}=0$ for $i=1,2,3$ and $j=4,5$. No, if you can't.
Sep
30
comment Covering Spaces in Representation Theory.
The full detail of this is in Assem-Simson-Skowronski's book IV.6.1 - IV.6.5. The main difficulty is to show that any projective object is of this form (which means the functor is dense); fully faithfulness of the functor is easy to see from Yoneda lemma. Note: a functor is fully faithful dense iff it gives an equivalence of categories.
Sep
16
comment Computing quotient representations and Hom set fort wo representations
Thanks! Now should be OK.
Sep
16
revised Computing quotient representations and Hom set fort wo representations
corrected edit
Sep
15
comment Computing quotient representations and Hom set fort wo representations
@Julian, thanks! This is embarrassing...Anyway, I have corrected the answer.
Sep
15
revised Computing quotient representations and Hom set fort wo representations
corrected calculation for char K not equals to 2
Sep
14
answered Computing quotient representations and Hom set fort wo representations
Sep
12
comment Direct sum decomposition of weight spaces and relation to Tensor products.
A small comment about answer to (Q3) as both answers didn't mention it. The (forumla for) decomposition of $V\otimes W$ for $V,W$ irreducible is usually called Clebsch-Gordan decomposition.
Jul
28
revised Can any $\theta \in \text{Hom}(S^\lambda,M^\mu)$ be written as $\theta = \kappa_t$?
corrected typeset