User aston smythe - MathOverflowmost recent 30 from http://mathoverflow.net2013-06-19T20:12:37Zhttp://mathoverflow.net/feeds/user/1977http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/27827/genus-of-complex-projective-spaceGenus of complex projective spaceAston Smythe2010-06-11T15:44:57Z2010-06-22T20:33:13Z
<p>The complex projective line is isomorphic to the 2-sphere, and so, has genus $0$. Does this result for all $CP^N$, that is, is the genus of $CP^N$ equal to $0$, for all $N$?</p>
http://mathoverflow.net/questions/27839/genus-of-grassmannians-and-flag-manifoldsGenus of Grassmannians and Flag Manifolds Aston Smythe2010-06-11T17:09:19Z2010-06-12T06:49:56Z
<p>In light of the answers given to <a href="http://mathoverflow.net/questions/27827/genus-of-complex-projective-space" rel="nofollow">this question</a>, I would like to pose a more general one: Do all Grassmannian spaces have genus 0? If so, do there exist any flag manifolds with non-zero genus? </p>
http://mathoverflow.net/questions/19219/smooth-manifolds-that-dont-admit-a-partition-of-unity Smooth manifolds that don't admit a partition of unityAston Smythe2010-03-24T18:07:52Z2010-03-24T21:41:17Z
<p>When I first starting studying differential geometry, I asked my lecturer a question about smooth manifolds that didn't admit a partition of unity. He promptly told not to worry about such objects as they were only studied by the extremely eccentric. I would like to know if this is true, ie, does anyone study manifolds that don't admit a partition of unity (not whether such people are eccentric).</p>
http://mathoverflow.net/questions/18948/invariant-vector-fields-for-homogenous-spacesInvariant Vector Fields for Homogenous SpacesAston Smythe2010-03-21T19:23:34Z2010-03-21T20:22:52Z
<p>As we all know, the space of invariant vector fields on a Lie group can be identified with the tangent space at the identity (or any other point for that matter). My question is: How does this generalize to homogeneous spaces? My guess would be that one can equate the space with the tangent space at any point point. However, it is not clear to me why everything should carry over smoothly to this more general setting.</p>
http://mathoverflow.net/questions/17666/what-to-call-the-elements-of-a-tensor-productWhat to call the elements of a tensor product.Aston Smythe2010-03-09T23:05:04Z2010-03-09T23:25:37Z
<p>What does one call the first (or second) factor of an element of a tensor product? For example, if $V,W$ are vector spaces, and $v \in V$, $w \in W$, with $v \otimes w \in V \otimes W$, how would one refer to $v$? First tensor factor? </p>
http://mathoverflow.net/questions/11821/su2-and-the-three-sphere$SU(2)$ and the three sphereAston Smythe2010-01-15T04:52:08Z2010-02-01T10:45:57Z
<p>Can anyone give me an explicit isomorphism between $SU(2)$ and the three sphere?</p>
<p>What about for higher spheres? This question <a href="http://mathoverflow.net/questions/11628/complex-projective-space-as-a-u1-quotient" rel="nofollow">link text</a> seems to indicate that there exists a homeomorphism from $SU(n)/SU(n-1)$ to the $(2n-1)$-sphere.</p>
http://mathoverflow.net/questions/13303/hodge-index-theorem-for-complex-manifoldsHodge Index theorem for Complex ManifoldsAston Smythe2010-01-29T00:10:09Z2010-01-29T02:42:17Z
<p>The Riemann-Roch theorem is a result about Riemann Surfaces that was extended to the Hirzebruch–Riemann–Roch theorem, a result about compact complex manifolds. The Hodge Index theorem is a result about Riemann surfaces (I'm just worried about the complex case) that is proved using Riemann-Roch. Has the Hirzebruch–Riemann–Roch theorem been used to extend the Hodge Index theorem to a result about compact complex manifolds.</p>
http://mathoverflow.net/questions/13002/principle-bundle-connection-correspondence-for-two-descriptions-of-the-mathbbcPrinciple Bundle Connection Correspondence for two descriptions of the $\mathbb{CP}^2$Aston Smythe2010-01-26T01:43:39Z2010-01-26T04:26:44Z
<p>Consider the following pair of principle bundle descriptions of $\mathbb{CP}^2$:
$$
\mathbb{CP}^2 \simeq SU(3)/U(2) \simeq S^5/U(1).
$$
If I have a principle $U(2)$-bundle connection for $\mathbb{CP}^2$, will that correspond to a principle $U(1)$-bundle connection? (Where by correspond I suppose I mean give the same covariant derivative.) </p>
<p>I am also interested in how this generalises to the $n$-case
$$
\mathbb{CP}^{n} \simeq SU(n+1)/U(n) \simeq S^{2n+1}/U(1).
$$</p>
http://mathoverflow.net/questions/11628/complex-projective-space-as-a-u1-quotientComplex Projective Space as a $U(1)$ quotientAston Smythe2010-01-13T06:10:22Z2010-01-13T19:56:57Z
<p>As is well known, one can view $\mathbb{CP}^n$ as a quotient of the unit $(2n + 1)$-sphere in $\mathbb{C}^{n+1}$ under the action of $U(1)$, since every line in $\mathbb{C}^{n+1}$ intersects the unit sphere in a circle. </p>
<p>Moreover, we have $S^{2n + 1} = SU(n+1)/SU(n)$, where $SU(n)$ embeds into the bottom right-hand corner (say).</p>
<p>My question is: Is there an embedding $j$ of $U(1)$ into $SU(n+1)/SU(n)$ that gives the $U(1)$ action as a left (right) multiplication, i.e. such that $A.e^{i \theta} = Aj(e^{i \theta})$, for all $A \in SU(n+1)/SU(n)$?</p>
<p>For $n=1$, it's easy: $SU(2) = S^3$, and we embed $e^{i\theta}$ as<br />
$\left(
\begin{array}{cc}
e^{i \theta} & 0 \\
0 & e^{-i \theta}
\end{array}
\right)$.</p>
http://mathoverflow.net/questions/10724/relationship-between-algebraic-and-holomorphic-differential-formsRelationship between algebraic and holomorphic differential formsAston Smythe2010-01-04T17:48:10Z2010-01-04T19:02:03Z
<p>I'm a little confused and in need of some clarification about the
relationship between algebraic and holomorphic differential forms:</p>
<p>(1) What is the exact definition of the module of differential
forms of a complex projective variety?</p>
<p>(2) What is the definition of its differential?</p>
<p>(3) Am I right in assuming that the algebraic forms are a
submodule of the holomorphic forms with the two differentials
coinciding in some obvious sense?</p>
http://mathoverflow.net/questions/8285/badiou-and-mathematicsBadiou and MathematicsAston Smythe2009-12-09T01:25:21Z2009-12-09T14:07:33Z
<p>Does anyone have an opinion on Alain Badiou's use of set theory? Is there anything interesting mathematically there? Also could anyone shed any light on the comment in the Wikipedia article <a href="http://en.wikipedia.org/wiki/Alain%5FBadiou" rel="nofollow">link text</a> that says: </p>
<blockquote>
<p>This effort leads him, in Being and Event, to combine rigorous mathematical formulae with his readings of poets such as Mallarmé and Hölderlin and religious thinkers such as Pascal.</p>
</blockquote>
http://mathoverflow.net/questions/7439/algebraic-varieties-which-are-also-manifoldsAlgebraic Varieties which are also ManifoldsAston Smythe2009-12-01T14:58:56Z2009-12-01T17:28:02Z
<p>Any non-singular projective variety over $\mathbb{C}$ is easily seen to be a smooth manifold. Presumably the same is not true for algebraic varieties - one would not expect varieties with singular points to have a smooth structure. But do there exist non-singular varieties that are not smooth manifolds?</p>
http://mathoverflow.net/questions/6332/the-2-sphere-and-mathbbcp1The 2-sphere and $\mathbb{CP}^1$Aston Smythe2009-11-21T00:15:02Z2009-11-21T00:28:42Z
<p>As is very well known, the algebraic variety $S^2$ is isomorphic to projective variety $\mathbb{CP}^1$ as a complex manifold. As is also well known, the coordinate ring of $S^2$ is given by $< x,y,z > / < x^2 + y^2 +z^2 - 1 >$ and the function field of $\mathbb{CP}^1$ is $\mathbb{C}(\mathbb{CP}^1)$ (its coordinate ring being $\mathbb{C}$). My question is how the coordinate ring of $S^2$ and the function field of $\mathbb{CP}^1$ are related? Presumably this relation is a special case of a general variety-complex manifold relationship.</p>
http://mathoverflow.net/questions/29547/interesting-consequences-of-the-riemann-roch-formula-for-line-bundles-over-flag-mComment by Aston SmytheAston Smythe2010-06-25T21:13:45Z2010-06-25T21:13:45ZYes. But I prefer to think about it as a special case of the Hirzebruch-Riemann-Roch formula.
http://mathoverflow.net/questions/27839/genus-of-grassmannians-and-flag-manifolds/27840#27840Comment by Aston SmytheAston Smythe2010-06-11T17:24:13Z2010-06-11T17:24:13Z... and the arithmetic genus?
http://mathoverflow.net/questions/23627/group-and-hopf-algebra-structures-for-projective-varietiesComment by Aston SmytheAston Smythe2010-05-05T22:39:17Z2010-05-05T22:39:17ZI suppose I mean when its coordinate ring when we consider it as a real affine variety - so scrap projective from the question and just consider affine.
http://mathoverflow.net/questions/19219/smooth-manifolds-that-dont-admit-a-partition-of-unityComment by Aston SmytheAston Smythe2010-03-24T18:29:52Z2010-03-24T18:29:52ZI mean manifolds that aren't second countable but are locally euclidean and Hausdorff .... do we need to specify Hausdorff here, surely it should from the manifold being locally euclidean.http://mathoverflow.net/questions/18948/invariant-vector-fields-for-homogenous-spaces/18951#18951Comment by Aston SmytheAston Smythe2010-03-22T20:44:40Z2010-03-22T20:44:40ZWell, as a grad student I really like "Grad student/homework" answers. http://mathoverflow.net/questions/18948/invariant-vector-fields-for-homogenous-spaces/18951#18951Comment by Aston SmytheAston Smythe2010-03-22T12:27:39Z2010-03-22T12:27:39ZCan one construct a basis of all vector fields using the invariant ones, like in the Lie group case?http://mathoverflow.net/questions/17666/what-to-call-the-elements-of-a-tensor-productComment by Aston SmytheAston Smythe2010-03-09T23:18:24Z2010-03-09T23:18:24Z@ Andrea: But I've got a collection of about 116 $v \otimes w$'s, and I really don't want to refer to them individually. I need a way to say "Take all the first tensor factors and consider the ideal they generate".http://mathoverflow.net/questions/13632/transition-functions-and-complex-structure/13677#13677Comment by Aston SmytheAston Smythe2010-02-02T00:31:13Z2010-02-02T00:31:13Z ... and a similar situation holds for $\Omega^{(1,0)}(M)$ and $\Omega^{(0,1)}(M)$?
http://mathoverflow.net/questions/13632/transition-functions-and-complex-structure/13677#13677Comment by Aston SmytheAston Smythe2010-02-01T23:05:17Z2010-02-01T23:05:17ZTo see if I understand I am going to take the example of $\mathbb{CP}^1$. The change-of-coordinate map
from $\phi_1(U_1) \subset \mathbb{C}$ to $\phi_2(U_2) \subset \mathbb{C}$ is $z \mapsto z^{-1}$. The complex Jacobian of this map is $z \mapsto -z^{-2}$. This gives the bundle $\mathcal{O}(2)$? The conjugate is $z \mapsto \overline{z}^{-1}$, which is not $\mathcal{O}(-2)$ as I thought it should be.http://mathoverflow.net/questions/13303/hodge-index-theorem-for-complex-manifolds/13312#13312Comment by Aston SmytheAston Smythe2010-01-29T14:33:51Z2010-01-29T14:33:51ZHow does this relate to versions that involve the rank of the Neron-Severini group?http://mathoverflow.net/questions/13002/principle-bundle-connection-correspondence-for-two-descriptions-of-the-mathbbc/13009#13009Comment by Aston SmytheAston Smythe2010-01-26T23:39:47Z2010-01-26T23:39:47ZI am I right in assuming then that any $U(1)$-connection induces a $U(2)$ connection via the embedding $\mathfrak{u(1)} \to\mathfrak{u(2)}$? http://mathoverflow.net/questions/11821/su2-and-the-three-sphereComment by Aston SmytheAston Smythe2010-01-15T17:41:54Z2010-01-15T17:41:54ZYes, of course, I just mean homeomorphisn (or diffeomorphism if you want to get differential).http://mathoverflow.net/questions/11628/complex-projective-space-as-a-u1-quotient/11651#11651Comment by Aston SmytheAston Smythe2010-01-13T17:44:46Z2010-01-13T17:44:46Z@Andrew David's answer is what I was looking for. I knew that $\mathbb{CP}^n$ was a $U(1)$ quotient of the $(2n + 1)$-sphere because of the relationship between planes and the shpere explained above, but I wanted a concrete matrix version of the action.http://mathoverflow.net/questions/11628/complex-projective-space-as-a-u1-quotientComment by Aston SmytheAston Smythe2010-01-13T06:47:26Z2010-01-13T06:47:26ZI mean an embedding such that the quotient of $SU(n+1)/SU(n)$ under the resulting multplicative action of $U(1)$ is homeomorphic to $\mathbb{CP}^n$.http://mathoverflow.net/questions/10724/relationship-between-algebraic-and-holomorphic-differential-forms/10725#10725Comment by Aston SmytheAston Smythe2010-01-04T18:33:02Z2010-01-04T18:33:02Z@jvp So, just to be sure, you're saying that for projective smooth complex varieties: algebraic and holomorphic differential forms coincide?