18,989 reputation
23779
bio website utm.edu/staff/jdevito
location Martin, TN
age 30
visits member for 4 years, 4 months
seen 9 hours ago

I'm an assistant professor at the University of Tennessee at Martin, in my fourth semester here.

My research interests include compact Riemannian manifolds with positive/non-negative sectional curvature, and, in particular, those with a "large" isometry group. I love working with compact Lie groups, especially homogeneous spaces and biquotients.

At some point, I intend to learn some set theory.


1d
awarded  Good Question
Nov
20
comment Short proof of Borsuk-Ulam's
I don't mind writing it (but won't have time for a bit), but if you'd like to write the answer yourself, I also don't mind that.
Nov
19
comment Short proof of Borsuk-Ulam's
I don't see a better way of making this work than the "usual" way. In the usual way, $g$ (defined on $S^n$), is restricted to $S^{n-1}$ and then its argued that the degree $g|_{S^{n-1}}:S^{n-1}\rightarrow S^{n-1}$ must be odd. Once you have this fact, this implies that $g$ (defined on $\mathbb{R}P^n$), when restricted to $\mathbb{R}P^{n-1}$ is nonzero on $H^{n-1}$ with $\mathbb{Z}_2$ coefficients. The cup product structure then shows that $g^\ast$ on $H^1$ is nontrivial, a contradiction.
Nov
19
comment Short proof of Borsuk-Ulam's
I've gotta run off to class, but I find homology much easier to think about. Note that, since $\mathbb{Z}/2\mathbb{Z}$ is a field, $H_1 = H^1$ in a natural sense. So, if you can show $g_\ast$ is nonzero on $H_1$, that also gives a contradiction.
Nov
19
answered Examples of group homomorphisms with isomorphic but not equal images
Nov
18
comment showing an inner product space isn't a bounded operator
Compute $( T(\alpha_n), T(\alpha_n) )$ and compare it with $(\alpha_n, \alpha_n)$ to show that no $K$ works uniformly.
Nov
13
comment Smooth structure on $M\cup_f N$?
@PtF: Then I think my comment is a counter example to your question. In fact, I think that most of the time $M\cup_f N$ is not a manifold at all. (In fact, I don't know of a single situation, other than $U = V = M = N$ where the $M\cup_f N$ has the structure of a manifold at all....
Nov
13
comment Smooth structure on $M\cup_f N$?
Just drawing pictures, it's not clear to me that $M\cup_f N$ is a manifold at all. For example, if you do this where $M = N = S^1$, don't you get something homeomorphic to the number $8$?
Nov
12
comment What is the “opposite” of the Axiom of Choice?
@Asaf: Thank you!
Nov
12
comment What is the “opposite” of the Axiom of Choice?
@Asaf: Logic is definitely not my strong suit (nor my weakest), but what does your sentence "there is a theorem that the axiom of choice implies the law of excluded middle" mean? More precisely, how can an adoption of a particular sentence $(AC)$ as an axiom, in whatever language you're working with, impact your choice of logical axioms?
Nov
12
comment Intersection of finite measure coverings of the rationals?
I think you can use the Baire category theorem. There may be an easier way
Nov
11
comment Spaces of constant curvature
Not that I know of....
Nov
11
comment Spaces of constant curvature
Ah, this is proved in more or less every Riemannian geometry book. For example, it's in Do Carmo's book on page 163. It's on page 145 of Petersen's book. Is this what you want?
Nov
11
comment Spaces of constant curvature
As stated, this is false: $\mathbb{R}P^n$ also has an isotropic metric of constant curvature. I'm pretty sure that this is the only exception. (It definitely is in the positive curvature setting).
Nov
7
comment A function from a smooth manifold with boundary to $[0,\infty)$
Do you already know that $\partial M$ has a neighborhood diffeomorphic to $\partial M \times [0,1)$?
Nov
6
comment Is there only one free (continuous) action of $\mathbb{Z}_2$ on $S^2$?
@studiosus: I had some vague idea that Perelman's theorem implied that a closed $3$-manifold with finite fundamental group was a quotient of $S^3$, but I didn't realize one could assume the action of the deck group is conjugate to a subgroup of $O(4)$. Granting that, I agree it's simple linear algebra from there.
Nov
4
comment Relation between Map and Dimension
a surjective map from $\mathbb{R}^n$ to whatever manifold you want. The composition is a surjective continuous map from $\mathbb{R}$ to whatever manifold you want. (You're right that a compact manifold can't surject onto a non-compact manifold.)
Nov
4
comment Relation between Map and Dimension
I think there is still a surjective continuous map with domain $\mathbb{R}$. For example, there is a surjective continuous map from $\mathbb{R}$ to $\mathbb{R}^2$, obtained by iterating the usual Peano curve. There is an easy surjective map $\mathbb{R}^2\rightarrow S^2$ given by wrapping a plane around $S^2$ infinitely many times. The composition gives a surjective map from $\mathbb{R}$ to $S^2$. (More generally, one can iterate this to find a surjective continuous map $\mathbb{R}\rightarrow \mathbb{R}^n$. Then, after choosing a complete Riemannian metric, the exponential map gives
Nov
4
comment The properness of a submersion
Your counterexample is also a counterexample to my proposed solution. I'll have to keep thinking about it.
Nov
3
comment What is the importance of Metric in Riemannian Geometry?
@studiousus: I didn't mean to imply you didn't know of Perelman's work - I intended that statement for other readers (sorry for not indicating that clearly). Thank you for the edit!