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Given a function $f(z)$ on the complex plane, define the Schwarzian derivative $S(f)$ to be the function $S(f) = \frac{f'''}{f'} - \frac{3}{2} \Big(\frac{f''}{f'}\Big)^2$ Here is a somewhat more conce …

I am not an expert, but there seems to be an enormous technical difference between algebraic geometry and differential/metric geometry stemming from the fact that there is apparently no such thing as …

A couple years ago Atiyah published a claimed proof that $S^6$ has no complex structure. I've heard murmurs and rumors that there are problems with the argument, but just a couple months ago he appar …

It seems that nature "is" a Klein bottle in the following sense. There is a growing field in applied topology (yes, I said that) which goes by "topological data analysis" or sometimes "persistent hom …

Recall that the scalar curvature of a Riemannian manifold is given by the trace of the Ricci curvature tensor. I will now summarize everything that I know about scalar curvature in three sentences: …

Atiyah and Hirzebruch gave a rather dramatic answer to your question in their paper "Spin Manifolds and Group Actions": if $M$ is a compact smooth spin manifold of dimension $4k$ whose $\hat{A}$-genus …

As far as I know, this sort of structure was first invoked by Dirac in order to take a square root of the Laplacian, and this he was doing in order to write down Lorentz invariant Klein-Gordon equatio …

Yikes, that's brutal - Kobayashi-Nomizu is an excellent reference text, but using it in a first course on the subject is a bit like learning English from the Oxford English Dictionary. For instance: …

A well known theorem of Cartan states that every free homotopy class of closed paths in a compact Riemannian manifold is represented by a closed geodesic (theorem 2.2 of Do Carmo, chapter 12, for exam …

Here is an updated formulation of the question, which is more precise and I think completely correct: Suppose $M$ is a Riemannian manifold. Pick a point $p$ in $M$ and let $U$ be a neighborhood of th …

Here is my precise question. Let $M, \omega$ be a symplectic manifold and let $H: M \to \mathbb{R}$ be any smooth function. The symplectic form gives rise to an isomorphism between the tangent bundl …

My question is motivated by the following well-known computation. Let $M$ be an even dimensional Riemannian spin manifold and let $D$ be the spinor Dirac operator on $M$. Lichnerowicz showed that $D …

I don't have any magical references for you, nor do I understand the NCG point of view on the Penrose tiling all that well. I learned just enough about this to convince myself that I didn't need to l …

1) The eta invariant itself depends on the metric, but the relative eta invariant is in many cases (see comments) a homotopy invariant. The relative eta invariant is defined to be the difference of t …

I agree with @coudy's answer that the best approach is to first understand the theorem's special cases / applications / generalizations. That can help highlight some of the key pain points in the var …