Complex Projective Space as a $U(1)$ quotient - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T23:27:53Z http://mathoverflow.net/feeds/question/11628 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/11628/complex-projective-space-as-a-u1-quotient Complex Projective Space as a $U(1)$ quotient Aston Smythe 2010-01-13T06:10:22Z 2010-01-13T19:56:57Z <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 /> $\left( \begin{array}{cc} e^{i \theta} &amp; 0 \\ 0 &amp; e^{-i \theta} \end{array} \right)$.</p> http://mathoverflow.net/questions/11628/complex-projective-space-as-a-u1-quotient/11635#11635 Answer by S. Carnahan for Complex Projective Space as a $U(1)$ quotient S. Carnahan 2010-01-13T06:49:38Z 2010-01-13T06:49:38Z <p>From wikipedia: $\mathbb{CP}^n$ is a <a href="http://en.wikipedia.org/wiki/Symmetric%5Fspace#Classification%5Fresult" rel="nofollow">Symmetric space of type AIII</a> for $p=n$, $q=1$. There are embeddings of both $U(1)$ and $SU(n)$ into $S(U(n) \times U(1)) \subset SU(n+1)$ which give you your quotient by right multiplication.</p> http://mathoverflow.net/questions/11628/complex-projective-space-as-a-u1-quotient/11640#11640 Answer by David Bar Moshe for Complex Projective Space as a $U(1)$ quotient David Bar Moshe 2010-01-13T07:44:22Z 2010-01-13T07:44:22Z <p>The U(1) group is the torus in SU(N+1) which commutes with SU(N). In the example given in the question where SU(N) is chosen as the bottom N-dimensional block. U(1) consists of the diagonal matrices </p> <p>diag{exp(N*i*theta), exp(-i*theta), . . . . (N-times) exp(-i*theta)}</p> <p>Please observe that the restriction to the bottom N-dimensional block is proportional to the unit matrix thus it commutes with the whole of SU(N), also it belongs to SU(N+1), since its has a unit determinant.</p> <p>The reason that the U(1) and the SU(N) factors commute is due to a theorem by <a href="http://www.pnas.org/content/40/12/1147.full.pdf+html" rel="nofollow">A. Borel</a> which states that the denominator subgroup of homogeneous Kaehlerian spaces must be the centralizer of a torus. In our case the torus is the U(1) subgroup and the certralizer is SU(N)*U(1)</p> http://mathoverflow.net/questions/11628/complex-projective-space-as-a-u1-quotient/11689#11689 Answer by Andrew Stacey for Complex Projective Space as a $U(1)$ quotient Andrew Stacey 2010-01-13T19:56:57Z 2010-01-13T19:56:57Z <p>My original answer was unsalvegable so I've deleted it and am posting a new "answer". As with the first one, I don't rate this as particularly an answer but more just trying to understand what's going on.</p> <p>I was initially having trouble understanding Scott's answer, but now I think I do and I think it gives the matrix representation wanted which isn't quite what David wrote.</p> <p>We have $SU(n+1)$ and inside this we have $SU(n)$ and quotient out to get $S^{2n+1}$. We also have a slightly larger subgroup which is $S(U(n) \times U(1))$, which contains $SU(n)$, such that the quotient is $\mathbb{CP}^n$.</p> <p>Now, $S(U(n) \times U(1))$ is $U(n)$ via $A \mapsto (\det A^{-1},A)$ and the inclusion $SU(n) \to S(U(n) \times U(1))$ goes over to the standard inclusion. Here, $SU(n)$ is a normal subgroup and $U(n)$ is the semi-direct product of $SU(n)$ and $U(1)$ with the map $U(1) \to U(n)$ given by $\lambda \mapsto (\lambda, 1,\dots,1)$ (diagonal matrix). When taken over to $S(U(n) \times U(1))$ this becomes $\lambda \mapsto (\lambda^{-1},\lambda,1,\dots,1)$.</p> <p>So then $SU(n+1)/S(U(n) \times U(1)) \cong (SU(n+1)/SU(n))/U(1)$ where $U(1) \to SU(n+1)$ is the map $\lambda \to (\lambda^{-1},\lambda,1,\dots,1)$.</p> <p>This isn't the same as David's, I know, so it may not be what you want (since that answer's been accepted). Presumably only one satisfies the condition that you want and presumably it's David's since that answer's been accepted. Still, I was confused and I think I've straightened myself out now.</p>