1,962 reputation
213
bio website
location Norway
age 23
visits member for 1 year, 11 months
seen 16 hours ago

I'm a 2st 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.


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 $\dim \mathscr L(\mathbb R^m,\mathbb R^n;\mathbb R^p)$
Sep
13
comment $\dim \mathscr L(\mathbb R^m,\mathbb R^n;\mathbb R^p)$
Do you know what a tensor product is?
Sep
5
comment The space of collars of a manifold is contractible
Okay, I think I understand. If we require the collars to intersect $\partial M$ transversely, then at least for small enough values for $t$, the pullback into the other collar will define the graph of a function, right? It should go like $(f(x,t),r(x)t)$ for some positive real-valued function r(x). In my head it should be like the proof using morse functions of the existence of collar neighborhoods.
Sep
5
comment The space of collars of a manifold is contractible
Why is it a problem? I don't see why I cannot do this. If I can scale the second map to land in the image of the first, by scaling it sufficiently its preimage through the first map will have connected fibers over $\partial M$, so we should be able to scale it up again (or shrink $e_0$) to match them. What goes wrong?
Sep
5
revised The space of collars of a manifold is contractible
edited tags
Sep
5
asked The space of collars of a manifold is contractible
Sep
5
answered The space of collars of a manifold is contractible
Aug
21
answered The “Circle” is a Vector Space?
Jul
2
awarded  Curious
Jun
5
accepted Degree 3 algebraic curve with a triple point
Jun
4
comment Degree 3 algebraic curve with a triple point
That makes perfect sense! Thank you very much!
Jun
4
comment Degree 3 algebraic curve with a triple point
@TedShifrin Do you mean that $V(f)$ is non-reduced as an affine scheme?
Jun
4
asked Degree 3 algebraic curve with a triple point
May
31
awarded  Nice Answer