s.b

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seen Oct 9 '12 at 3:52

Sep
12
awarded  Yearling
Oct
6
comment Bessel functions with complex coefficients
Mathematica of course has what yo want ... but you probably do not want it. I understand that Bessel Function is a function of exponential decay...so I am wondering if computing the Taylor series to few hundred terms (with Stirling approximation for the Gamma function) is not an option ?
Oct
6
comment Are there any decompositions of a symmetric matrix that allow for the inversion of any submatrix?
Assuming your matrix is real matrix (all elements are real), you are guaranteed by spectral theorem that your matrix $A = QDQ^{T}$ where $D$ is diagonal and $Q$ is orthogonal. Getting this $Q$ is algorithmically Gram-Schimdt process (so rather simple).
Sep
21
comment A basic question on relative singular groups\homology
Example $H_{1} \subset G_{1}$ and $H_{2} \subset G_{2}$ are two groups and their subgroups and $\phi : G_{1} \rightarrow G_{2}$ is any homomorphism taking $H_{1}$ into $H_{2}$ then we get a well defined homomorphism (assuming$H_{1}$ and $H_{2}$ are both normal; which is true in abelian case) of the groups $G_{1}/H_{1} \rightarrow G_{2}/H_{2}$.
Sep
21
revised Complex tori as elliptic curves
It is Silvenrman's book not Singerman.
Sep
21
suggested suggested edit on Complex tori as elliptic curves
Sep
21
comment A basic question on relative singular groups\homology
Yes you are right the boundary map descends to the quotient nicely for any subspace $A$ but the associated homology theory will not be nice unless some extra conditions on $A$ are fulfilled.
Sep
20
comment Methods of Multilinear Algebra in Representation Theory
The study of Lie algebras over $\mathbb{C}$ becomes easy once you have understood $\mathfrak{sl}_{2}$ and it might be a great idea to work out symmetric and tensor products of representation in this case by hand. Fulton Harris does that and if I remember correctly they have a beautiful (and very verbose) discussion on that topic.
Sep
20
comment Methods of Multilinear Algebra in Representation Theory
Thanks! for the link
Sep
20
answered A basic question on relative singular groups\homology
Sep
20
answered Methods of Multilinear Algebra in Representation Theory
Sep
20
answered Representation of finitely generated groups
Sep
18
comment Segre embedding
This map is a closed embedding into $\mathbb{P}^{rs+r+s}$ which means that the image is a closed algebraic subscheme (or rather subvariety). To show this show that in any affine patch of $\mathbb{P}^{rs+r+s}$ the points of the Segre embedding correspond to an ideal. Writing co-ordinates for $\mathbb{P}^{rs+r+s}$ as $k[X_{ij}]$ where $0\leq i\leq r$ and $0 \leq j \leq s$ we see that the image of the Segre embedding correspond to the ideal generated by all $2 \times 2$ minors of the matrix $(X_{ij})$. This shows that the image is closed.
Sep
17
comment Is it possible to practice mental math too often?
You might be hitting the so called "ok plateau".
Sep
16
comment Proving something by copying down axioms and changing variable names?
You need to assert that an axiom is true in your case of this set of even functions. Then check that your assertion is true.
Sep
16
comment (Symmetric) group acting on a graph
Here is an attempt: Let $f: X \rightarrow Y$ be a surjective map of graphs (vertices surject onto vertices and so do edges; basically it is a contraction of certain edges and vertices)then we can talk about the group Aut(X/Y) which are graph automorphisms of $X$ commuting with the map $f$. In your case group $P$ is the group $Aut(\tilde{X}/(X/G))$. Well here $\tilde{X}=$ universal cover of $X$ and $X/G$ the quotient graph of $X$ by $G$.
Sep
15
comment homeomorphism question relating to the topological 3-sphere
@MarianoSuárez-Alvarez, I recall when I was first learning this stuff, this question stumped me for some time. Where as when you look at any sphere (after learning some basic general topology) the idea of breaking it up into two hem-spheres seems much more natural.
Sep
15
comment Is the set of integer coefficient polynomials countable?
@Thomas Andrews, Yes I meant the polynomial algebra not a single polynomial per se.
Sep
14
comment Is the set of integer coefficient polynomials countable?
An integral polynomial gives a function from $\mathbb{Z} \rightarrow \mathbb{Z}$. A power series doesn't. A polynomial is a finitely generated algebra over the co-efficient ring a power series is not.
Sep
14
comment Set of Bounded linear Operators on $l_2$ is dense on the set of bounded operators on $l_2$?
A hlibert space is still a vector space...what do you mean by a operator which is not linear?