I was trying to find any information about the first couple of homotopy groups of the complex Grassmannian $G_\mathbb{C} (2,4)$ of complex planes in complex $4$-space. I need the first seven or eight homotopy groups.
Best Regards,
Stephan
What are the first 8 homotopy groups of the complex Grassmannian $G_\mathbb{C}(2,4)$?
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algebraic-topology
grassmannian
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1Why do you need them? – 2017-02-10
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1It is part of some approach to discretise $G_\mathbb{C}(2,4)$ as a simplicial set, and then a simplicial complex. The idea is to start with an existing model of $\mathbb{CP}^2$ and attach the remaining three cells starting from the top dimensional one. We have a recent result which will show me how to attach the cells if I am able to find which homotopy classes of $\mathbb{CP}^2$ are killed by the attaching maps. The first thing that came to my mind was to simply compare the homotopy groups of $G_\mathbb{C}(2,4)$ and $\mathbb{CP}^2$ and see if I could figure out which classes are killed. – 2017-02-10
1 Answers
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This is more of a comment than an answer, but it's not too hard to compute the first four homotopy groups. In particular, there is a well-known fiber bundle $$ U(2) \;\longrightarrow\; V_{\mathbb{C}}(2,4) \;\longrightarrow\; G_{\mathbb{C}}(2,4), $$ where $$ V_{\mathbb{C}}(2,4) \,=\, \bigl\{(v,w)\in\mathbb{C}^4\times\mathbb{C}^4 \;\bigl|\; \|v\|=\|w\|=1\text{ and }\langle v,w\rangle = 0\bigr\} $$ is the corresponding Stiefel manifold. Using the octonians, it is easy to show that $$ V_{\mathbb{C}}(2,4) \,\approx\, S^7 \times S^5. $$ In paricular $\pi_n\bigl(V_{\mathbb{C}}(2,4)\bigr) = 0$ for $n\leq 4$. Using the long exact sequence for a fibration, it follows that $$ \pi_n\bigl(G_{\mathbb{C}}(2,4)\bigr) \,\cong\, \pi_{n-1}\bigl(U(2)\bigr) $$ for $n\leq 4$. The homotopy groups for $U(2)$ can be computed using the fiber bundle $S^1\to U(2)\to S^3$; in particular $\pi_1\bigl(U(2)\bigr)\cong\mathbb{Z}$ and $\pi_n\bigl(U(2)\bigr)\cong \pi_n(S^3)$ for all $n\geq 2$. Therefore, $$ \pi_4\bigl(G_{\mathbb{C}}(2,4)\bigr)\cong \mathbb{Z},\quad \pi_3\bigl(G_{\mathbb{C}}(2,4)\bigr)= 0,\quad \pi_2\bigl(G_{\mathbb{C}}(2,4)\bigr)\cong \mathbb{Z},\quad\text{and}\quad \pi_1\bigl(G_{\mathbb{C}}(2,4)\bigr)=0. $$ The long exact sequence doesn't seem sufficient to compute the homotopy groups for $n\geq 5$, so I'm not sure how to push this any further. For example, the fifth homotopy group fits into a short exact sequence $\mathbb{Z} \to \pi_5\bigl(G_{\mathbb{C}}(2,4)\bigr) \to \mathbb{Z}_2$, but I'm not sure how to tell whether it's $\mathbb{Z}$ or $\mathbb{Z}\times\mathbb{Z}_2$.
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0Dear Jim, I don't see the above short exact sequence. The the according portion of the long exact sequence looks like: $\dots \rightarrow \mathbb{Z}_2 \rightarrow \mathbb{Z} \rightarrow \pi_5(G_{\mathbb{C}}(2,4)) \rightarrow \mathbb{Z}_2 \rightarrow 0$ , because $\pi_5(U(2)) \cong \pi_5(\mathbb{S}^3) \cong \mathbb{Z}_2$. – 2017-02-13
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0@StephanZhechev The only homomorphism from $\mathbb{Z}_2$ to $\mathbb{Z}$ is the zero homomorphism, so you can separate the long exact sequence at this point. – 2017-02-13
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0Yes, of course. Sorry for the stupid comment. – 2017-02-13
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0After some consideration I figured out what I wanted and your answer was very helpful, so I accept it and close the problem. Thank you. – 2017-02-28