bio | website | |
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location | Austin, TX | |
age | 28 | |
visits | member for | 4 years, 6 months |
seen | yesterday | |
stats | profile views | 1,074 |
I recently finished my PhD at UPenn under the supervision of Jonathan Block. I will be starting a post-doc at UT Austin in the Fall.
Jul 26 |
revised |
Complex structures on Riemann surfaces
added 169 characters in body |
Jul 25 |
comment |
Complex structures on Riemann surfaces
Thanks @JasonStarr, I've fixed it. |
Jul 25 |
revised |
Complex structures on Riemann surfaces
added 4 characters in body |
Jul 25 |
asked | Complex structures on Riemann surfaces |
Jul 24 |
answered | A Lie algebra identity |
Jul 2 |
awarded | Curious |
Jun 27 |
comment |
Integrating representations of Lie algebroids
Thanks for the nice answer! |
Jun 27 |
accepted | Integrating representations of Lie algebroids |
Jun 26 |
asked | Integrating representations of Lie algebroids |
May 12 |
awarded | Favorite Question |
Apr 21 |
awarded | Enlightened |
Apr 21 |
awarded | Nice Answer |
Apr 20 |
comment |
If there exists a nontrivial vector field $V$ such that $\nabla_{X}V=0$ for any vector field $X$, the manifold must be flat?
I should say that above $n = \dim M$. |
Apr 20 |
comment |
If there exists a nontrivial vector field $V$ such that $\nabla_{X}V=0$ for any vector field $X$, the manifold must be flat?
Say $V_1,\ldots,V_n$ is a collection of linearly independent (everywhere on $U$) vector fields that are parallel, i.e. $\nabla_X V_j = 0$ for all vector fields $X$. For any vector fields $X,Y$, $R(X,Y)$ is an endomorphism of the tangent bundle, where $R$ is the Riemann curvature tensor. Since it is an endomorphism (as opposed to a differential operator), to show it is zero it suffices to show it vanishes on a frame of the tangent bundle. But $R(X,Y) V_i = \nabla_X \nabla_Y V_i -\nabla_Y\nabla_X V_i - \nabla_{[X,Y]} V_i = 0$ for all $i$. |
Apr 20 |
awarded | Informed |
Apr 20 |
answered | If there exists a nontrivial vector field $V$ such that $\nabla_{X}V=0$ for any vector field $X$, the manifold must be flat? |
Mar 30 |
comment |
Atiyah classes of holomorphic vector bundles with trivial Chern classes
@PavelSafronov Err, now I'm confused. Topologically we have $\mathcal (O(-1)\otimes \mathcal O(-1))\oplus \mathbb C = \mathcal O(-1) \oplus \mathcal O(-1)$. So topologically, $\mathcal O(1) \oplus \mathcal O(-1) = \mathbb C^2$ is trivial and so admits a flat connection. |
Mar 30 |
comment |
Atiyah classes of holomorphic vector bundles with trivial Chern classes
@PavelSafronov Thanks. Also I just realized since $\pi_1(\mathbb P^1) = 0$ the only flat rank 2 vector bundle is trivial. |
Mar 30 |
accepted | Atiyah classes of holomorphic vector bundles with trivial Chern classes |
Mar 30 |
comment |
Atiyah classes of holomorphic vector bundles with trivial Chern classes
Thanks for the answer! Do you happen to have a reference for the theorem of Weil? Also, do you know offhand if $\mathcal{O}_{\mathbb{P}^1}(p)\oplus \mathcal{O}_{\mathbb{P}^1}(-p)$ admits a flat connection? |