2,077 reputation
315
bio website
location Norway
age 23
visits member for 2 years, 1 month
seen Nov 21 at 23:55

I'm a 2nd year master's student of mathematics, specializing in algebraic topology.

I have Bc.S. degrees in mathematics and physics.

My interests include topology, geometry and mathematical physics, and more specifically abstract homotopy theory and cobordism theory.


Nov
21
comment Power series modulo polynomials
@user26857 I may do that once I have had time to work a bit more on it, in light of the comments I recieved here.
Nov
16
reviewed Reject suggested edit on Points of Inflection
Nov
16
awarded  Custodian
Nov
16
reviewed Approve suggested edit on Finding the zeroes in a function
Nov
15
revised Lesser known derivations of well-known formulas and theorems
added 16 characters in body
Nov
15
answered Lesser known derivations of well-known formulas and theorems
Nov
14
comment Power series modulo polynomials
Thank you for the references! So the question can be summarized as: "Is an abelian group always a pure subgroup of its completion wrt. a separated filtration?" right?
Nov
14
comment Power series modulo polynomials
I see. There is a problem with infinite sums, of course. Thank you for pointing it out. Does the failure to be pure exact persist if we replace Z with a (alg. closed?) field? Also, in what way is it nasty?
Nov
14
asked Power series modulo polynomials
Nov
2
answered What is the Quotient on the Coproduct in Adjunction Spaces
Oct
29
comment A conjecture concerning primes and algebra
You can probably post this on Mathoverflow to get a higher chance of an answer.
Oct
24
awarded  Yearling
Sep
24
awarded  Autobiographer
Sep
14
comment Question regarding adjoint functors
For the best result, you want to respect the natural abelian group structure on the Hom-sets, so it should be a group homomorphism. The natural choice of isomorphism will give you a $B$-module isomorphism though.
Sep
13
comment $\dim \mathscr L(\mathbb R^m,\mathbb R^n;\mathbb R^p)$
Choose bases $x_1,...,x_n$ of $\mathbb{R}^n$, $y_1,...,y_m$ of $\mathbb{R}^m$ and $z_1,...,z_p$ of $\mathbb{R}^p$. The bilinear map $f_{ijk}$ takes the basis vector pair $x_i,y_j$ to the basis vector $z_k$ of $\mathbb{R}^p$ and all other basis vector pairs to the zero vector. I.e. $f_{ijk}(x_a,y_b) = z_k$ if $a=i$ and $b=j$, and zero otherwise.
Sep
13
comment $\dim \mathscr L(\mathbb R^m,\mathbb R^n;\mathbb R^p)$
I tried to give a more explicit explanation now.
Sep
13
revised $\dim \mathscr L(\mathbb R^m,\mathbb R^n;\mathbb R^p)$
added 602 characters in body
Sep
13
answered Where to study $2$-category theory?
Sep
13
comment $\dim \mathscr L(\mathbb R^m,\mathbb R^n;\mathbb R^p)$
That's fine. A tensor product simply encodes a "universal" bilinear map, but you don't need it for this problem.
Sep
13
answered $\dim \mathscr L(\mathbb R^m,\mathbb R^n;\mathbb R^p)$