Eric O. Korman

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	age	27
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Graduate student in mathematics.

	7,515 reputation	bio	website		visits	member for	2 years, 10 months
1731 badges		location	New Jersey		seen	11 hours ago

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Apr 24	answered	Positive curvature on holomorphic vector bundles
Apr 24	comment	What does $\Omega^\bullet(M)$ mean? There really is no difference. $\bullet$ and $$ are often used interchangeably when denoting $\mathbb Z$-graded objects. $\bullet$ may be the better notation though since $$ is also used for dual.
Apr 24	comment	Basic questions to a principal $S^1$ bundle over M @gofvonx: that is not a group action. Let $g = e^{i\pi}$. Then $p \cdot g = p + \pi$ so $(p \cdot g) \cdot g = p + 2\pi$. But $g^2$ is the identity element of $S^1$ so you must have $p \cdot g^2 = p$. This problem is simply wrong unless $\mathbb R$ is really supposed to be $S^1$.
Apr 24	comment	Every principal $G$-bundle over a surface is trivial if $G$ is compact and simply connected: reference? @DaanMichiels: The main thing about classifying spaces to know is that the universal $G$-bundle $EG$ over $BG$ is contractible. Then the long exact sequence in homotopy associated to the fibration $G \to EG \to BG$ tells you that $\pi_n(BG) = \pi_{n-1}(G)$.
Apr 24	answered	What does $\Omega^\bullet(M)$ mean?
Apr 24	awarded	differential-geometry
Apr 23	answered	Classification of flat complex line bundles
Apr 23	revised	Classification of flat complex line bundles edited tags
Apr 23	asked	Classification of flat complex line bundles
Apr 22	comment	Orthogonal group is a regular submanifold of $GL(n,\Bbb R)$ You can also consider $f$ as mapping into the vector space of symmetric matrices. Then you can show very explicitly that $D_{A_0} f$ is surjective.
Apr 21	comment	An abelian group of finite order has an element of the order of the group. This is only true for cyclic groups and not all abelian groups are cyclic.
Apr 21	revised	Bing's House and homotopies added 187 characters in body
Apr 21	comment	Bing's House and homotopies @DonAntonio so this question is asking about the inclusion of $S^1$ in a cone?
Apr 21	comment	Bing's House and homotopies ¿quien es Bing?
Apr 19	comment	Computing the homology groups of a given surface @ShaiDeshe Actually now I'm not so sure how clear it is that $U \cap V$ is a retract of $V$, but this argument should be made to work.
Apr 19	comment	Computing the homology groups of a given surface To kernel of $\mathbb Z \to H_1(\Delta^2/\sim)$ is the image of the isomorphism $\mathbb Z \to \mathbb Z$. So the map $\mathbb Z \to H_1(\Delta^2/\sim)$ is just the zero map. But this map is also surjective because of exactness, so $H_1(\Delta^2/\sim) = 0$.
Apr 19	answered	Computing the homology groups of a given surface
Apr 18	awarded	Good Answer
Apr 18	comment	Basic questions to a principal $S^1$ bundle over M Also, I don't think there is a clear (non-trivial) action of $S^1$ on $\mathbb R$. Maybe there is a typo somewhere in your homework set?
Apr 18	comment	Basic questions to a principal $S^1$ bundle over M For a principal bundle, the action should also be transitive (which it isn't in your case since the fibers of the projection aren't $S^1$).