17,177 reputation
23276
bio website utm.edu/staff/jdevito
location Martin, TN
age 30
visits member for 3 years, 11 months
seen 5 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.


Jul
18
awarded  Sportsmanship
Jul
18
answered Find a CW complex with prescribed homology groups
Jul
16
comment R-linear functionals on manifolds
I should point out that this is a standard style of proving that tensorial things only depend on points, and not on neighborhoods.
Jul
16
answered R-linear functionals on manifolds
Jul
13
comment Question on unitary representation of non-compact simple Lie groups
Why is the map $\mathfrak{g}\rightarrow \mathfrak{su}(n)$ surjective?
Jul
9
comment How to show that $f_* (\sigma)=\sigma$ where $f$ is mapping between projective spaces $\mathbb{R}\text{P}^3$
@user110822: I'm glad you like this answer. I'll just leave it as is - if someone needs to see the proof, I'll add it or answer it as a separate question, or whatever.
Jul
8
answered How to show that $f_* (\sigma)=\sigma$ where $f$ is mapping between projective spaces $\mathbb{R}\text{P}^3$
Jul
6
comment Transfinite Induction and the Axiom of Choice
@Trevor: For (1), I didn't mean "meta" in the usual sense logicians use "meta", so perhaps a better word should replace it. I meant it more in the sense of "unraveling induction" - nothing formal about it. For (2), I agree that it looks misleading, but I hoped the damage was repaired via the statement "Of course, the failure of this approach doesn't mean there is NO way of proving choice from $ZF$..." Perhaps I should clarify it. For the benefit of the OP (or anyone else), I believe that Trevor's answer is a much better answer to the question. I'd suggest accepting his instead of mine.
Jul
2
awarded  Curious
Jun
29
comment A question on the unit tangent bundle of the sphere and $SO(3)$
@Qiaochu: I figured you'd object to my map as "canonical" because of the choice of basis ;-). I, personally, am of the mindset that $\mathbb{R}^n$ comes with a God-given orthonormal basis (and to me, the notation $SO(3)$ means the orthogonal group relative to the usual inner product on $\mathbb{R}^3$), so I still view my map as "canonical". Now, if we were, say, talking about $V$ being the degree at most 2 polynomials over $\mathbb{R}$ with $L^2$ inner product, I'd agree that my map is non-canonical. Also, you're right compactness is an essential ingredient - thanks!
Jun
28
comment A question on the unit tangent bundle of the sphere and $SO(3)$
@Peter: It never is diffeo to a Lie group for $n>2$. The point is that $T^1 S^n$ is simply connected, so if a Lie group, it must be semisimple. But all semisimple Lie groups have $H^3$ nontrivial, even rationally. This immediately implies, since $T^1 S^n$ is $n-1$ connected, that $n\leq 4$. When $n=3$, $T^1 S^3 = S^3 \times S^2$ has nontrivial $\pi_2$, so can't be Lie group. When $n=4$, $T^1 S^4$ is rationally $6$-connected.
Jun
28
comment A question on the unit tangent bundle of the sphere and $SO(3)$
@Qiaochu: There is, in fact, a canonical diffeomorphism (for some definition of canonical): Map $(p,v)\in T^1 S^2$ to the matrix with columns $p,v,$ and $p\times v$. (And +1 from me)
Jun
25
answered Are there more embeddings $U(2) \hookrightarrow SO(4)$?
Jun
25
awarded  Revival
Jun
25
answered What are necessary and sufficient conditions for the product of spheres to be paralellizable?
Jun
16
comment Total dimension of the cohomology of a homogeneous space (or of a graded Tor)
@jdc: (In case you care, I do agree with all the edits you made. Thanks!)
Jun
12
comment 'Obvious' theorems that are actually false
@Lost1: It actually doesn't depend on the field: $\vec{v}\cdot \vec{0} = 0$ and $\{\vec{v}, \vec{0}\}$ is never independent.
May
29
comment Total dimension of the cohomology of a homogeneous space (or of a graded Tor)
to compute the normalizer of a subgroup in general. In fact, this very problem has been causing me some difficulties lately in my own research.
May
29
comment Total dimension of the cohomology of a homogeneous space (or of a graded Tor)
I see now that I typed $BT_H$ in agreement with your previous post, but in my head I was definitely thinking of $BT_G$. In general it should be easy to work out for subtori of classical groups, simply because the inclusion $i:T\rightarrow G$ factors through the inclusions $T\rightarrow T_G\rightarrow G$. Thus, the map on cohomology also factors. The first map is the one $H^\ast(BG)\rightarrow H^\ast(BT_G)$ alluded to above. The second $H^\ast(BT_G)\rightarrow H^\ast(BT)$ is computed in the same way I did in the example post. Also, to answer the question in your email, no, I don't know how
May
28
comment Total dimension of the cohomology of a homogeneous space (or of a graded Tor)
(First, I agree with your edit - sorry about that!) In general, I don't have a good understanding of the map $H^\ast(BG)\rightarrow H^\ast(BT_H)$. But for the classical groups, it's fairly well understood (completely with $\mathbb{Q}$-coefficients, and also for $\mathbb{Z}$-coefficients if you ignore the SO groups.)