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seen Sep 30 at 15:33

Dec
5
awarded  Yearling
Sep
30
awarded  Explainer
May
17
answered Why is abstract algebra so important?
May
17
comment Series of functions as Lebesgue integral with counting measure
you are integrating with respect to $n$, so you want to view $f_{n}(x)=: f(x,n)$ as a function of $n$.
May
16
comment Can one prove the existence of tensor product without explicitly constructing it?
I also said, I suspect. I am not certain it will work.
May
16
comment Can one prove the existence of tensor product without explicitly constructing it?
@Qiaochu Yuan Not true, there is a more general version which is applicable in any model category
May
16
answered Can one prove the existence of tensor product without explicitly constructing it?
May
16
answered Intuition and Motivation - Linear Operator $T - \lambda_k I$ ? [Lay P270 Thm 5.1.2]
May
16
comment Convergent subsequence proof
looks good to me
May
10
comment Question about the Fundamental group of circle.
Qiaochu is correct. Imagine your loop $\alpha_{1}$ is a string with both ends nailed to a table (one nail: both ends are in the same spot). Now imagine that $\alpha_{3}$ is a string which is looped around itself 3 times, then its endpoints are nailed. No matter how you move or stretch the first string you will not be able to make it look like the second. This is precisely because you are not allowed to move the endpoints.
May
10
comment Let $\alpha= f\,dx_1 \wedge \cdots\wedge dx_n$; where $f$ is continuous on $A$. Show that $\int_ \Phi \alpha =\int_ \Phi f$
I think you really want $\int_{A}f$ where the integral is taken in the measure theoretic sense. The LHS of the equation should refer to evaluation of the cochain $\alpha$ on the chain $\Phi$.
May
1
comment Unitary operator on dense set, Unique extension?
If you have a bounded extension $\tilde{U}$ then certainly it is the unique bounded operator extending $U$. This just follows from a general topological fact: if two continuous functions agree on a dense set, then they agree everywhere.
Apr
30
comment Maximum value of line integral
Isn't this just a standard variational problem? I would parametrize the curve by $(x(t),y(t))$ and use the Euler-Lagrange equations.
Apr
30
comment Soft: Why does the existence of a singularity cause problems for deRham cohmology?
If you really want to know exactly what breaks down, it is the face that smooth differential forms cannot be defined. In fact, a basic starting point to defining the forms is that the tangent space at every point is isomorphic. Smooth differential $k$ forms are defined to be $\Lambda^{k}(V)\otimes C^{\infty}(M;\mathbb{R})$, where $V$ is a typical fiber of the tangent bundle. However, at a singular point, the tangent space is not well defined. Neither are smooth functions.
Apr
30
comment Soft: Why does the existence of a singularity cause problems for deRham cohmology?
Well, one issue is that locally, around a singular, one cannot define a smooth patch. For example, any self intersecting curve has a singular point at the self intersection. Around this point, the variety looks something like a cross, So it does not locally "look" like the real line.
Feb
16
comment Stalk of the quotient presheaf
I agree, this is the way to approach the problem. I might add that, since you are working on the level of presheaves, you need the added fact that sheafification preserves these colimits (this is essentially what was pointed out below). This is, of course, if you want to get the result for the corresponding sheafification of the quotient.
Jan
8
answered How to imagine a fractal dimension?
Jan
5
comment How to determine vector space?
For a subset of a vector space one only needs to check these closure axioms since any subset inherits the operations of the vector space.
Jan
5
comment How to determine vector space?
@hardmath functions on finite sets can be identified with vectors in a finite dimensional space.
Jan
5
answered Example of continuous function that isn't uniformly continuous and isn't 1/x