19,519 reputation
23983
bio website utm.edu/staff/jdevito
location Martin, TN
age 30
visits member for 4 years, 6 months
seen 19 mins 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
comment Prove that $az^n+b\overline{z}^n=0$ does not have any complex solutions except for $0$
Since you're given a condition on $|a|$ and $|b|$, try rearranging the question and then taking the magnitude/norm/length of both sides.
2d
comment Is there a way to define the concept of manifolds so it looks more like “generalised affine spaces”?
Also, in case it helps anyone's intuition, on a parallelizable Riemannian manifold $M$, one can use $\exp:M\times T_p M\rightarrow M$ as $\boxplus$. I'm still not sure what to make of $\boxminus$.
2d
comment Is there a way to define the concept of manifolds so it looks more like “generalised affine spaces”?
I don't have anything concrete to say, but my instinct is that this may work for parallelizable manifolds, but probably doesn't other wise. For whatever reason, I feel like your definition of $\boxminus$ is forcing that. In any case, it feels as though your definitions are too global in character to replace the usual notion of manifold, which is inherently local in character. Again, I have nothing concrete now, so I don't yet have counter example. Have you tried working out the $S^2$ case in detail?
2d
comment Surprising applications of topology
It may also be worth mentioning that there is a complete characterization (in terms of presentations) of which finite groups can act freely on $S^n$. This is only hard when $n = 4k-1$: When $n$ is even, only $\mathbb{Z}_2$ can act freely, and if $n = 4k+1$, only cyclic groups act freely. (Actually, I'm not 100% positive about the $4k+1$ case. Certainly all finite cyclic groups act freely, but conversely...?)
2d
comment Surprising applications of topology
Did you mean $S^3$ everywhere you wrote $S^2$?
Jan
27
comment Finding $\mathbf{10}\otimes \mathbf{8}\otimes \mathbf{8}\otimes \mathbf{8}$ in $SU(3)$
Not to my knowledge. The only notation I've ever used is by labeling the nodes in the Dynkin diagram by natural numbers, or variations on that.
Jan
26
comment Simplifying a direct sum $\mathbf{3}\oplus\mathbf{3}\oplus\mathbf{2}$ etc
You can simplify it just as you did, but you have to be very careful that you keep bold/nonbold stuff straight. Also, your original calculation could have simplified to $2(\mathbf{10}\otimes \mathbf{8})$.
Jan
26
comment Finding $\mathbf{10}\otimes \mathbf{8}\otimes \mathbf{8}\otimes \mathbf{8}$ in $SU(3)$
Well, I kind of cheated. I don't know your notation well enough to say which one is barred, so in my head, I defined $X[4,1] = 35$. In terms of the Dynkin indices, it then follows that $X[1,4] = \overline{35}$. (I'm actually heading to lunch, then class, so I won't be able to respond for another 4 hours or so...)
Jan
26
comment Finding $\mathbf{10}\otimes \mathbf{8}\otimes \mathbf{8}\otimes \mathbf{8}$ in $SU(3)$
@LoveLearning: I wish I had a nice answer. I used the website quite a bit about 6-7 years ago when trying to figure out some stuff for my thesis, and I remember a lot of how it worked from then. The $A_2 \iff SU(3)$ notation is fairly standard (see classification of simple Lie algebras), and the recognition that, e.g., $X[3,0]\iff 10$ came from plugging things into their "dimension of a module" option, and noting that dimension $10$ only occurred for $X[3,0]$ and $X[0,3]$.
Jan
25
comment What is the modern axiomatization of (Euclidean) plane geometry?
@Paul: It's literally been years since I've thought about this. If I recall correctly, Hilbert's axioms fix everything up. I'm not sure which "minimal" axiom of Hilbert's would fix this particular issue, though.
Jan
24
comment Finding $\mathbf{10}\otimes \mathbf{8}\otimes \mathbf{8}\otimes \mathbf{8}$ in $SU(3)$
To answer your comment, yes, $10\otimes 1 = 10$. Also, I'm not sure what size inputs the program I linked to can handle (I think it can only handle single digit highest weights).
Jan
24
answered Finding $\mathbf{10}\otimes \mathbf{8}\otimes \mathbf{8}\otimes \mathbf{8}$ in $SU(3)$
Jan
24
comment Finding $\mathbf{10}\otimes \mathbf{8}\otimes \mathbf{8}\otimes \mathbf{8}$ in $SU(3)$
Yes, tensor products of reps distribute over sums of reps, so (1) is ok. I'm not sure how, e.g. $10\otimes 27$ decomposes.
Jan
21
comment Nonintegrable almost complex structures
@Michael: It's probably also worth adding that my only interaction with (almost) complex manifolds has been through a few cursory google searches - I've probably spent a grand total of a couple of days of my life thinking about them in any kind of detail. Sorry to be useless!
Jan
21
comment Nonintegrable almost complex structures
@Michael: I see. In retrospect, I guess I don't know that it's "due to the non-associativity of the octonions". That said, I wrote that because in my head, the only algebraic difference between the quaternions and octonions is associativity, so it "must" be the reason. I suppose a truly accurate phrasing would be "I think this complex structure is non-integrable, because I know that finding an integrable structure is a famous unsolved problem. I suspect that the non-integrability stems from the non-associativity of the octonions, but I'm really just speculating."
Jan
21
comment Nonintegrable almost complex structures
@Michael: Actually I don't. The "almost" part is pretty easy to prove. The fact that is is not integrable follows from the fact that, as t.b. said, "the existence of an integrable almost complex structure on $S^6$ is a famous open problem." According to this MO link: mathoverflow.net/questions/1973/…, LeBrun showed that there is no orthogonal complex structure on $S^6$, but the above almost complex structure is orthogonal.
Jan
21
awarded  riemannian-geometry
Jan
20
answered Three linearly independent vector fields
Jan
20
comment Three linearly independent vector fields
Since there are no non-vanishing continuous vector fields on $S^2$, there must be a point where $X$ is the $0$ vector, and hence $(0,X) = (0,0)$ at that point, so you no longer have independent vector fields.
Jan
16
comment Are there algebraic subgroups of Lie groups which happen to be Lie groups on their own but not Lie subgroups?
Glad you like it! Also, as in the second answer here: math.stackexchange.com/questions/565552/… one can see that, as a group, $S^1$ is isomorphic to $S^1\times \mathbb{R}$. Then one can use the argument in my answer to create discontinuous functions with domain $S^1$. This shows that "semisimple" part of van der Waerden's theorem is necessary.