10,749 reputation
32677
bio website math.stanford.edu/~jmadnick
location Stanford, CA
age 24
visits member for 4 years, 1 month
seen 8 hours ago

Third-year graduate student at Stanford University. My interests originate from differential geometry. My advisors are Robert Bryant and Rick Schoen.

My favorite aspects of math are those which appear to generalize calculus. I like the algebraic and analytic machinery of geometry and topology.

I learn best from concrete examples, explicit calculations, and relating modern ideas to their classical counterparts.

Currently learning about:

  • Riemannian manifolds with special holonomy
  • Gauge Theory
  • Calibrated geometries
  • Exterior differential systems
  • Geometric analysis & PDE

Sep
10
reviewed Approve suggested edit on Why are there infinitely many connections on a Riemannian manifold?
Sep
8
comment A curve internally tangent to a sphere of radius $R$ has curvature at least $1/R$ at the point of tangency
However, $\langle a,b \rangle + \langle b,a\rangle = \langle a,b \rangle + \langle a,b\rangle = 2\langle a,b\rangle$, if that helps. Anyway, that's all for me. Good luck!
Sep
8
comment A curve internally tangent to a sphere of radius $R$ has curvature at least $1/R$ at the point of tangency
@Unkle: $\langle a,b \rangle + \langle c,d \rangle \neq \langle a+c, b+d \rangle$.
Sep
8
comment A curve internally tangent to a sphere of radius $R$ has curvature at least $1/R$ at the point of tangency
@Unkle: Your second derivative $f''(s_0)$ looks odd to me. For instance, why does the inner product have three terms? If $f(s) = \langle a(s), a(s) \rangle$, then $$f'(s) = \langle a'(s), a(s) \rangle + \langle a(s), a'(s) \rangle,$$ yes?
Sep
7
revised A curve internally tangent to a sphere of radius $R$ has curvature at least $1/R$ at the point of tangency
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Sep
7
answered A curve internally tangent to a sphere of radius $R$ has curvature at least $1/R$ at the point of tangency
Sep
7
comment A curve internally tangent to a sphere of radius $R$ has curvature at least $1/R$ at the point of tangency
Sorry, I should have been clearer. I mean that you have not proven why $\Vert a''(s_0) \Vert \geq \Vert \gamma''(s_0) \Vert$ holds. Your reason why it ought to hold geometrically is persuasive, but I think one can be much more precise.
Sep
7
comment A curve internally tangent to a sphere of radius $R$ has curvature at least $1/R$ at the point of tangency
The "therefore it is not too hard to see that" is precisely what the OP is trying to prove. What you have given is not a proof, but an intuitive explanation as to why the fact in question ought to be true.
Sep
5
comment Is there a function whose derivative is $|x|$?
Somewhat minor point: The last line shows that $\lim_{x \to 0} f'(x) = 0$. This is different from showing that $f'(0)$ exists and equals $0$. In other words, you haven't shown that $f(x) = \frac{x|x|}{2}$ is differentiable at $x = 0$.
Sep
5
comment Categorising types of Mathematics
Analysis
Aug
7
awarded  Nice Answer
Aug
4
awarded  Yearling
Jul
26
comment Flat connection with non-trivial holonomy? I cannot get it
Where did you come across these facts?
Jul
6
awarded  Nice Answer
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jun
17
comment Chirality of a Möbius band without boundary?
@studiosus: I'm aware. I was just trying to help clarify the OP's question, both so he could better understand it, and so that answerers would have a better idea what exactly he's asking (and where he's coming from / what mis-conceptions he may have).
Jun
17
comment Chirality of a Möbius band without boundary?
Let's see if I have this right. The OP is assuming some sort of correspondence between the Mobius strip and the real projective plane. Since the Mobius strip can be created either with a "left-handed twist" or "right-handed twist," the OP reasons that there are two different "versions" of the Mobius strip. Question 1 asks how these two different versions manifest themselves in the real projective plane -- and in particular whether there are in fact two "versions" of the real projective plane. @Gerard, is this what you're after?
Jun
14
comment Is every scalar differential operator on $(M,g)$ that commutes with isometries a polynomial of the Laplacian?
@WillJagy: What about it? Are you citing that as a counter-example to the fact I cited above [Folland, "Real Analysis: Modern Techinques and their Applications", Theorem 8.51]?
Jun
14
revised Why is the Laplacian important in Riemannian geometry?
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