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Homotopy $\pi_4(SU(2))=\mathbb{Z}_2$

+ 9 like - 0 dislike
751 views

Recently I read a paper using $$\pi_4(SU(2))=\mathbb{Z}_2.$$ Do you have any visualization or explanation of this result?

More generally, how do physicists understand or calculate high dimension homotopy group?


This post imported from StackExchange Physics at 2014-03-22 17:01 (UCT), posted by SE-user Yingfei Gu

asked Dec 8, 2012 in Mathematics by Yingfei Gu (105 points) [ revision history ]
retagged Mar 25, 2014 by dimension10
Most voted comments show all comments
<google.com/search?q=Π4(S3)>

This post imported from StackExchange at 2014-03-22 17:01 (UCT), posted by SE-user Francois Ziegler
$SU(2)$ is homeomorphic to $S^3$ so you are asking why is $\pi_4(S^3) = \mathbb{Z}/2$? Pontryagin came up with a method to do such computations using bordisms of framed submanifolds; see thm 21 on page 99 of math.rochester.edu/people/faculty/doug/otherpapers/pont4.pdf.

This post imported from StackExchange at 2014-03-22 17:01 (UCT), posted by SE-user solbap
Actually use $\pi_4$, not $\Pi_4$, which is a different construction. Also, $\pi_4(S^2) = \pi_4(S^3)$.

This post imported from StackExchange at 2014-03-22 17:01 (UCT), posted by SE-user David Roberts
@David what is $\Pi_4$?

This post imported from StackExchange at 2014-03-22 17:01 (UCT), posted by SE-user Mariano Suárez-Alvarez
The generator of $\pi_{4}(S^{3})$ is the image of the Hopf map under the suspension homomorphism $\Sigma: \pi_{3}(S^{2}) \rightarrow \pi_{4}(S^{3})$. There are a few cool ways to visualize the Hopf map; try Googling.

This post imported from StackExchange at 2014-03-22 17:01 (UCT), posted by SE-user Sam Nolen
Most recent comments show all comments
@Alexander, got it. Thanks for reminding.

This post imported from StackExchange at 2014-03-22 17:01 (UCT), posted by SE-user Yingfei Gu
Cross post: mathoverflow.net/questions/115866/homotopy-pi-4su2z-2

This post imported from StackExchange Physics at 2014-03-22 17:01 (UCT), posted by SE-user student

5 Answers

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Since $SU(2)$ is topologically a three-sphere $S^3$, you can begin by investigating the homotopy groups of spheres. Unfortunately, although there are some regular results, such as $\Pi_n(S^n)=\mathbb{Z}$, and $\Pi_m(S^n)=0$ for $m<n$, I don't think there is a single method to calculate $\Pi_m(S^n)$ for $m>n$. Individual results for $m>n$ are chaotic. So, I think the answer to your last question is "they would ask a mathematician!", because this (algebraic topology) is a very large topic.

For your specific case, there is a reference given on math overflow, but I don't have the book unfortunately.

This post imported from StackExchange Physics at 2014-03-22 17:01 (UCT), posted by SE-user twistor59
answered Dec 8, 2012 by twistor59 (2,490 points) [ no revision ]
Sorry, I had not seen your answer. +1

This post imported from StackExchange Physics at 2014-03-22 17:01 (UCT), posted by SE-user Heidar
Thanks for nice reference. There seems having explanation by suspension of SU(2), I am trying to imagine it.

This post imported from StackExchange Physics at 2014-03-22 17:01 (UCT), posted by SE-user Yingfei Gu
Yes, section 4 of this ref describes the process. I don't fully understand but you start with the Hopf fibration $S^3 \rightarrow S^1$ ($S^3$ is an $S1$ bundle over $S^2$), and then apply suspension operation to increase the dimensionalities of the spheres.

This post imported from StackExchange Physics at 2014-03-22 17:01 (UCT), posted by SE-user twistor59
+ 6 like - 0 dislike

It is very hard to visualize these homotopy classes, since they correspond to maps $S^4\rightarrow SU(2)\approx S^3$. The homotopy groups of spheres (and any other space) are typically very difficult to calculate in generality and physicists typically ask mathematicians. But there exist simple results in the so-called "stable range" where there is a regular structure (Bott periodicity, $\pi_k(U(N)) = \pi_{k+2}(U(N))$ for large enough $N$), and there exist tools to calculate homotopy groups of certain spaces, such as the long exact sequence of a fibration.

For the case of spheres see the table on wikipedia, where the chaotic behavior is clear and $\pi_4(S^3)$ is listed. There is a very good review by Mermin (1) where you can learn how to visualize and calculate simple homotopy groups.

(1): Rev. Mod. Phys. 51, 591–648 (1979)

This post imported from StackExchange Physics at 2014-03-22 17:01 (UCT), posted by SE-user Heidar
answered Dec 8, 2012 by Heidar (855 points) [ no revision ]
+1 too. At least we agree it's hard!

This post imported from StackExchange Physics at 2014-03-22 17:01 (UCT), posted by SE-user twistor59
Thanks! I've read Mermin's RMP paper before and did learn a lot, but for this problem, the fibration method seems not work, at least not in a apparent way.

This post imported from StackExchange Physics at 2014-03-22 17:01 (UCT), posted by SE-user Yingfei Gu
@Yingfei Gu, I naively thought one might use some sort of generalization of Hopf fibration (which let you calculate $\pi_3(S^2))$. But it seems that it is not possible to do this in an obvious way.

This post imported from StackExchange Physics at 2014-03-22 17:01 (UCT), posted by SE-user Heidar
+ 6 like - 0 dislike

This calculation of $\pi_4(S^3)$ is also obtained in the paper R. Brown and J.-L. Loday, Topology, 26 (1987) 311-334, and also available here. In that paper, $S^3$ is regarded as the double suspension $SS$ of the circle $S^1$, which is itself seen as an Eilenberg-Mac Lane space $K(\mathbb Z,1)$. We obtain in Proposition 4.10 a determination of $\pi_4$ of the double suspension $SS$ of a $K(G,1)$ for any group $G$ as the kernel of a morphism $G \tilde{\wedge} G \to G$ defined by the commutator map, where the group $G \tilde{\wedge} G$ is the quotient of the free group on the set $G \times G$ by a set of relations satisfied by commutators. Hence the result for $\pi_4 SK(G,1)$ is easy to calculate if $G$ is abelian: in fact in that case it is $G \otimes G$, the tensor product of abelian greoups, factored by the relations $g \otimes h + h \otimes g$ for all $g,h \in G$.

Part of the intuition behind this is that the suspension $SX$ of a space $X$ is regarded as the union $C^+X \cup C^- X$ of two cones with intersection $X$, and this union is one to which our van Kampen type theorem for squares of spaces can apply. In fact we are dealing with the triad $(SX; C^+X, C^-X) $, which is the union of two triads $(C^+X;C^+X,X)$, $(C^- X; C^-X,X)$, and the union of these two "trivial" triads creates something in dimension $3$. If $\pi_2 X=0$, this gives a complete determination of $\pi_3 SX$, and further work gives a result on the double suspension in the given case.

So the intuition is that in homotopy theory, identifications in low dimensions have high dimensional homotopy implications, and to cope with this for gluing purposes we need algebraic structures with structures in a range of dimensions. The hope is that someday such structures will have applications in physics!

This post imported from StackExchange at 2014-03-22 17:01 (UCT), posted by SE-user Ronnie Brown
answered Dec 9, 2012 by Ronnie Brown (120 points) [ no revision ]
I fixed the link to your paper.

This post imported from StackExchange at 2014-03-22 17:01 (UCT), posted by SE-user Akhil Mathew
@Akhil: thanks!

This post imported from StackExchange at 2014-03-22 17:01 (UCT), posted by SE-user Ronnie Brown
+ 6 like - 0 dislike

This is really just a handwavy but perhaps more "visual" description of Pontryagin's result as cited by solbap in the comments above. Though I've written a huge block of text, there are some reasonably concrete three-dimensional pictures that you can build up in your head in this case, but it does take quite a bit of practice.

First, I assume that you are familiar with Pontryagin's construction relating the homotopy classes of maps to the k-sphere with framed (co-)bordism classes of codimension k submanifolds.

Check out Milnor's book Topology from the Differentiable Viewpoint if you're not familiar with this. Because your user profile says that you are interested in condensed matter physics, I'll add that this idea is used in the case of $k=2$ to draw some nice pictures of "homotopies around defects" in this paper of Teo and Kane.

Warmup, $\pi_3(S^2)$

As a warmup, let's try to visualize homotopy classes of maps from $S^3$ to $S^2$, i.e. the situation of the Hopf fibration. Pontryagin's construction says that we should be looking at bordism classes of framed codimension 2 submanifolds in $S^3$. 3-2=1, so we should be looking at 1-dimensional submanifolds, i.e. links in $S^3$. Here we have framed links in $S^3$ which can be visualized by drawing each component of the link with another parallel copy that winds around it, much like a ribbon.

You should convince yourself that all components in these framed links can be merged together into a single unknot with some integer framing. Thus what matters ultimately is the classification of possible framings. Imagine taking a 2D slice of $S^3$ transverse to a point $p$ of the framed link and placing the point $p$ at the origin of that plane. Then the framing at that point is just a choice of the $x$- and $y$- axes (i.e. a 2-dimensional frame). As we carry this plane along the original unknot, this choice of axes can rotate in that plane and so the classification of framings is naturally an integer. You may check that the inverse image of the North pole of the Hopf fibration is an unknot, and the inverse image of any other point on the sphere is an unknot which is linked once with it. Finally, you should see how you can build up any other homotopy class from "adding" Hopf fibrations together by putting multiple copies of this framed unknot together (possibly with opposite orientations), which gives a visualization of the group structure on the set of homotopy classes.

In this way you get a visualization of $\pi_3(S^2)$ by means of some pictures of framed circles. I can't resist here adding a link to this paper of DeTurck et al which gives some beautiful illustrations and description of the homotopy classes of maps from $T^3$ to $S^2$ with this tool.

$\pi_4(S^3)$

Now, you are interested in the case of homotopy classes of maps from $S^4$ to $S^3$. In this case you are now looking at framed links in $S^4$. You can still arrange for the link to become a single framed unknot by a sequence of bordisms. However, the framing can no longer be drawn with simply just a single parallel knot. Consider taking a 3-dimensional slice transverse to a point $p$ on the link in $S^4$ and let us place $p$ at the origin of our $R^3$ that we sliced with. In $S^4$, the framing of the link yields a choice of a 3-dimensional frame in this $R^3$ slice. And just as the relevant topological invariant of the framing in $S^3$ was how this frame rotates as we travel along the $S^1$ corresponding to our link component, leading to an element of $\pi_1(S^1)$ (the winding number), in $S^4$, we must now track how this 3-d frame rotates as we follow the $S^1$ of the link component. But now we are considering a continuous loop of choices of 3-dimensional orientations, i.e. an element of $\pi_1(SO(3))$, which is well known to be $\mathbb{Z}/2\mathbb{Z}$.

With this key ingredient of the 3-dimensional framing, hopefully you can see that $\pi_4(S^3)=\pi_1(SO(3))=\mathbb{Z}/2\mathbb{Z}$.

This post imported from StackExchange at 2014-03-22 17:01 (UCT), posted by SE-user j.c.
answered Dec 10, 2012 by j.c. (260 points) [ no revision ]
Thank you! This is a very very nice answer~

This post imported from StackExchange at 2014-03-22 17:01 (UCT), posted by SE-user Yingfei Gu
Thanks, feel free to ask for clarification; it can be hard to describe the pictures without a chalkboard.

This post imported from StackExchange at 2014-03-22 17:01 (UCT), posted by SE-user j.c.
+ 6 like - 0 dislike

This question has been posted also at http://mathoverflow.net/questions/115866/homotopy-pi-4su2z-2 with both geometric and algebraic (that was mine!) type of answers. The geometric answers tell of Pontjyagin's method of constructing explicit representations of maps to spheres. The algebraic methods tells of the answer from a general theorem which gives some results on homotopy groups of complexes, and when the conditions under which it works are satisfies, gives detailed and calculable algebraic answers.

This post imported from StackExchange Physics at 2014-03-22 17:01 (UCT), posted by SE-user Ronnie Brown
answered Dec 10, 2012 by Ronnie Brown (120 points) [ no revision ]
Thank you for pointing out the answers from mathoverflow. They are very interesting!

This post imported from StackExchange Physics at 2014-03-22 17:01 (UCT), posted by SE-user Heidar

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