Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,928 questions , 1,396 unanswered
4,846 answers , 20,597 comments
1,470 users with positive rep
501 active unimported users
More ...

Wess-Zumino term topology and topology of 5-dimensional manifold

+ 3 like - 0 dislike
172 views

Suppose we have theory in $3+1$ dimensions with $G\simeq SU(N)\times SU(N)$ global symmetry which is broken down to $H \simeq SU(N)$. Corresponding goldstone boson fields $\epsilon_{a}(x)$ parametrize coordinates of coset $G/H \simeq SU(N)$, and their allowed values in spacetime $S_{4}\sim S_{3}\times S_{1}$ (here $S_{3}$ is spatial coordinates and $S_{1}$ is compact euclidean time) define points in coset $G/H$. Since $\pi_{4}(SU(N))$ is trivial, then we may maps $S_{3}\times S_{1}$ into one point, so that we may extend $\epsilon_{a}(x)$ to $\epsilon_{a}(x, s)$, where $s \in (0, 1)$, and $\epsilon_{a}(x, 0) \equiv \epsilon_{a}(x)$, $\epsilon_{a}(x, 1) = 0$, and the set $y = (x, s)$ defines 5-dimensional manifold, the bound of which is our spacetime.

The manifold for this problem can be given as $S_{3} \times D$, where $D$ defines 2-dimensional disc. I've read that this is right due to trivial homotopy group $\pi_{1}(SU(N)) = 0$. How to show this?

asked Feb 20, 2016 in Theoretical Physics by NAME_XXX (1,010 points) [ revision history ]

I guess you have to be careful about the meaning of $\sim$ in $S_4 \sim S_3 \times S_1$. The 4-sphere $S_4$ and the product of the 3-sphere $S_3$ by the circle $S_1$ are not the same topologically (they are not homeomorphic). Any map from $S_4$ to $SU(N)$ is homotopic to a constant because $\pi_4(SU(N))=0$ but it is not the case of $S_3 \times S_1$ as $\pi_3(SU(N)))=\mathbb{Z}$. So I am not sure what you mean.

2 Answers

+ 3 like - 0 dislike

As pointed out in the comments, you have to be careful about the distinction between using $S^4$ and $S^3 \times S^1$, since the latter allows for non-trivial topological maps to $SU(N)$. I'll give two answers, depending on which of the two you actually meant:

If you work with the spacetime $S^4$:

Then indeed all maps to $SU(N)$ are topologically the same, so you can continuously make the map constant. Mathematically this means you can extend a map $\epsilon: S^4 \to SU(N)$ to a map $\tilde \epsilon: S^4 \times D^1 \to SU(N)$ such that $\tilde \epsilon(x,0) = \epsilon(x)$ and $\tilde \epsilon(x,1) = 1$ (or zero, if we are rewriting it in terms of the Lie algebra, I suppose).

Note: $D^n$ is notation for the $n$-dimensional disk.

Your question then becomes:

How can we relate a map $S^4 \times D^1 \to SU(N)$ to a map $S^3 \times D^2 \to SU(N)$?

This is simple: simply puncture $S^4$ at the north and south pole. The manifold without these two points is basically $S^3 \times D^1$ (this becomes intuitive if you make a drawing in lower dimensions: puncturing $S^2$ at the poles gives you a cylinder $S^1 \times D^1$). Hence `forgetting' these two points we get $\boxed{S^4 \times D^1} \to (S^3 \times D^1) \times D^1 = \boxed{S^3 \times D^2}$.

Important: it does not work the other way around: not every map $S^3\times D^2 \to SU(N)$ defines a map $S^4 \times D^1 \to SU(N)$ ! This is because $\pi_3(SU(N)) = \mathbb Z$, such that the former (note that the topologically distinct maps from $S^3 \times D^2$ and from $S^3$ are the same since $D^2$ is contractible) has many different topological mappings, whereas the latter is always topologically trivial. So in that sense I think it is dangerous to just work with the manifold $S^3 \times D^2$, as it will give you too many non-trivial solutions in this case!

If you work with the spacetime $S^3\times S^1$:

In this case you can have topologically distinct maps $S^3 \times S^1 \to SU(N)$, so you cannot always continuously deform them to a constant. So in this case we can not define the $\tilde \epsilon$ as I did above (or you did in your post). Nevertheless, we can of course always trivially extend a map $S^3 \times S^1 \to SU(N)$ to a map $S^3 \times S^1 \times D^1 \to SU(N)$ (for example $\epsilon(x,s) = \epsilon(x)$).

Your question then becomes:

How can we relate a map $S^3 \times S^1 \times D^1 \to SU(N)$ to a map $S^3 \times D^2 \to SU(N)$?

This is again simple, but now for a different reason, and now we indeed have to use the fact that $\pi_1(SU(N)) = 0$. Since $S^3$ appears in both cases, it is sufficient to show it for a fixed point in $S^3$. In other words we show how a map $S^1\times D^1 \to SU(N)$ relates to a map $D^2 \to SU(N)$. The former is a map from a cylinder, the latter a map from a $2$-sphere. Clearly they are related if we can sensibly shrink each cylinder end to a point. But that is exactly what $\pi_1(SU(N)) = 0$ is telling us! In more detail: at each end of the cylinder, we effectively have a map $S^1 \to SU(N)$ which we can continuously shrink to a constant. Visually this means that if we put half a $2$-sphere as a cap on one of the cylinder ends, our map extends smoothly.

(The nice thing in this case is that the equivalence works in both ways!)

answered Feb 25, 2016 by Ruben Verresen (205 points) [ revision history ]

Thank you! But now I see unsolved for me problem of ambiquity of the manifold for the Wess-Zumino term. In the Witten paper "CURRENT ALGEBRA, BARYONS, AND QUARK CONFINEMENT" the manifold of WZ term is $S_{3}\times D_{2}$. In Weinberg's QFT Vol. 2, however, the manifold is 5-dimensional ball. Is there exist some theorem which makes the choice between these two manifolds indistinguishable?

+ 3 like - 0 dislike

Aha, based on a comment to my previous post I now better understand the question at hand. I still stand by what I wrote in my original post, but I now see it is not the best answer for the real question, so let me write my new answer here.

Allow me to first carefully state the question/context I will be addressing:

We are interested in the Wess-Zumino term in four dimensions. This is defined similar to the more well-known WZ term in two dimensions, so let me first repeat the logic. For a given map $g: S^2 \to SU(N)$, one notes that since $\pi_2(SU(N)) = 0$ we can consider the gradual/continuous deformation $g_s$ where $g_1 = g$ and $g_0 = 1$. Note that we can interpret this as a function $g_s: D^3 \to SU(N)$, where $s$ denotes the radius of the $3$-ball. So we have extended our function such that our original function lives on the boundary. One then defines the WZ term as $S_\textrm{WZ}[g]= \int_{D^3} \textrm{Tr} \left( g_s^{-1} d g_s \right)^3$ (there is some pre-factor I am ignoring). This is supposed to be a function of $g$, but we have used the object $g_s$. So for the WZ term to be well-defined, it should be independent of our above choice of $g_s$. Well, it turns out that it is independent up to $2\pi$, so $e^{iS_\textrm{WZ}[g]}$ is well-defined. To see this, take $g_s: D^3 \to SU(N)$ and $\tilde g_s: D^3 \to SU(N)$, of course both satisfying $g_1 = \tilde g_1 = g$. The latter means we can actually glue them together to a function $h_s: S^3\to SU(N)$. We then want to show that $\int_{S^3} \textrm{Tr} \left( h_s^{-1} d h_s \right)^3$ is an integer multiple of $2\pi$. The reason is basically because on a closed manifold one can show that this integral is measuring the topological winding number of the manifold around the target space, in this case $\pi_3(SU(N)) = \mathbb Z$.

So we see that a WZ term for a map into the group manifold is defined by first extending it to a map on a higher-dimensional manifold, then defining some integral of that map, and to then argue it is independent of the extension we used topological properties of our group. The question is then asking about the four-dimensional case, in particular how the different manifolds used are consistent with each other.

Answer:

Let me first rephrase the `simple' case I described above in a more topological way, since it will help in higher dimension and also mimicks a bit how Witten talks about it. The WZ term is actually defined by a certain three-form $\omega_\textrm{WZ} \in \Omega^3(SU(N))$ on $SU(N)$. One can pull this back under $g_s$ to get a three-form on $D^3$, which one can integrate to get $S_\textrm{WZ}$. But in the case of $h_s$, which is defined on the compact manifold $S^3$, there is an alternative nicer viewpoint: to a first approximation, one can intuitively think of the $k$-th homology group of a manifold as being the equivalence class of all $k$-dimensional submanifolds, with the equivalence up to cobordism (meaning two submanifolds define the same homology element if they can together form the boundary of a higher-dimensional manifold; for example two points in a space define the same zero-dimensional homology class if there is a line in our space connecting them). So our map $h_s:S^3 \to SU(N)$ defines a submanifold/class $[S^3]_h \in H_3(SU(N))$. Moreover, it turns out our WZ form $\omega_\textrm{WZ}$ is closed, which in terms of de Rham cohomology means it defines a cohomology element $[\omega_\textrm{WZ}] \in H^3(SU(N))$. Moreover, by definition cohomology elements eat homology elements, by which I mean one can evaluate $[\omega_\textrm{WZ}] \left(  [S^3]_h \right)$. In fact, this is an equivalent way of writing the WZ-term $S_\textrm{WZ}$ ! What does this give us? Well, now it is written in a manifestly topological notation. We see our map $h_s: S^3 \to SU(N)$ picks out an integer of $H^3(SU(N)) = \mathbb Z$. More exactly, the Wess-Zumino term is normalized by requiring that if we take a map $f:S^3 \to SU(N)$ corresponding to $1 \in \pi_3(SU(N))$, that then $[\omega_\textrm{WZ}] \left(  [S^3]_f \right) = 1$.

So alright, let's now suppose our original space-time is $S^4$, and so we start with a map $g: S^4 \to SU(N)$. Since $\pi_4(SU(N)) = 0$, we can use the same logic I just described and get $g_s: D^5 \to SU(N)$. Note that our original map $g$ lives on the boundary $\partial D^5 = S^4$. In ``Global aspects of current algebra'' (1983), Witten defines the WZ term by choosing a particular $5$-form $\omega_{WZ} $ on $SU(N)$. It's then the same story as above: $S_\textrm{WZ} = \int_{D^5} g_s^* (\omega_{WZ})$ (where `$g_s^*$' means `pulling back under this map'). Similarly, to show this is independent (up to some integer) of the choice of $g_s$, we want to see that $\int_{S^5} g_s^* (\omega_{WZ})$ is an integer. But this is again for the same reasons as above: since $S^5$ is a closed manifold it defines a homology element of $H_5(SU(N))$ and since $\omega$ is a closed form it defines a cohomology element in $H^5(SU(N))$, such that our integral becomes the topological $[\omega_\textrm{WZ}] \left(  [S^5]_h \right)$. Again, the Wess-Zumino term is normalized such that if we take $f: S^5 \to SU(N)$ corresponding to $1 \in \pi_5(SU(N))$, that $[\omega_\textrm{WZ}] \left(  [S^5]_f \right) = 1$.

Similarly we could have started with a spacetime $S^3 \times S^1$ and a map $g: S^3 \times S^1 \to SU(N)$. We can think of $S^3 \times S^1$ as associating a circle to every point in $S^3$. Write the coordinate of $S^3$ as $\vec n$ and of $S^1$ as $\phi$. For a fixed value of $\vec n$, think of gradually shrinking our circle to a point, creating the space $S^3 \times D^2$. We can extend our map to $g_s: S^3 \times D^2 \to SU(N)$ by gradually trivializing the map $g(\vec n, \phi)$ for a fixed value of $\vec n$ such that at the point of collapse $g_0(\vec n) = g(\vec n, 0)$. Note that this is possible by $\pi_1(SU(N)) = 0$. Note that our spacetime again lives on the boundary: $\partial(S^3 \times D^2) = S^3 \times S^1$. We again define $S_\textrm{WZ} = \int_{S^3 \times D^2} g_s^* (\omega_{WZ})$. Also, again, to prove it is well-defined we consider the glued map $h_s : S^3 \times S^2 \to SU(N)$. Again this is a five-dimensional closed manifold so defines an element of $H_5(SU(N))$ and we can rewrite $S_\textrm{WZ} =  [\omega_\textrm{WZ}] \left(  [S^3 \times S^2]_h \right)$.

But... is it clear that $S_\textrm{WZ} =  [\omega_\textrm{WZ}] \left(  [S^3 \times S^2]_h \right)$ is an integer? We don't have the luxury as before where we could normalize our term such that we get integers, because we have already normalized it such that $[\omega_\textrm{WZ}] \left(  [S^5]_f \right) = 1$ (where $f$ corresponded to $1 \in \pi_5(SU(N))$). In principle it is possible that for example $[\omega_\textrm{WZ}] \left(  [M^5]_i \right) = \frac{1}{2}$ for some five-dimensional manifold embedded by $i: M^5 \to SU(N)$. Indeed, Witten mentions on p434 of ``Current algebra, baryons and quark confinement'' that there are indeed such examples! But it turns out that this can't happen if $M^5 = S^3 \times S^2$ :) To see this, we have to prove that if $f:S^3 \times S^2 \to SU(N)$ defines an element $[S^3 \times S^2]_f$ that we can find a map $g: S^5 \times SU(N)$ such that $[S^3 \times S^2]_f = [S^5]_g$. Now if you scroll up and see how I introduced homology above, you see that all we need to show is that there is a manifold $M$ whose boundary $\partial M$ is $S^3 \times S^2$ and $S^5$ (this is called a cobordism between these two manifolds) and a map $k:M \to SU(N)$ such that on the boundary $k$ reduces to $f$ and $g$. As I was writing this I thought I had an argument for this, but it didn't work out. I am pretty sure there must be a way of seeing it, but it's pretty late here and I should call it a day. I will try and complete the argument soon, but I also welcome any pointers. Witten on p434 implies all you need is $\pi_2(SU(3)) = 0$.

So that at least shows things are consistent: the WZ term is well-defined, even if we take our spacetime to be $S^3 \times S^1$. But I feel part of your question is also: can we get different physical phenomena depending on these two spacetimes? And that should indeed be possible: a $S^3\times S^1$ spacetime allows for non-trivial configurations in the far past (i.e. non-zero global soliton number), whereas a $S^4$ spacetime assumes everything becomes constant in the far past/future. So yes the physics can in principle be different, but the above shows how everything is well-defined in either case. I have tried to clarify the relationship between different kinds of topological information and defining the WZ term. As a quick recap: to define the WZ term starting from two dimensions and spacetime $\boxed{S^2}$, you use $\boxed{\pi_2(SU(N))= 0}$ to extend the map (with the auxiliary space $\boxed{D^3}$). To define the WZ term starting from a $\boxed{S^4}$ spacetime, you use $\boxed{\pi_4(SU(N))= 0}$ to extend the map (with the auxiliary space $\boxed{D^5}$). And if you start from $\boxed{S^3 \times S^1}$ you use $\boxed{\pi_1(SU(N))=0}$ to get a map on the auxiliary space $\boxed{S^3 \times D^2}$. Moreover we saw how one needs to check that the last two definitions are consistent/well-defined (they are). Lastly there is the comment that in principle the physics can depend on the choice of relevant spacetime.

answered Feb 28, 2016 by Ruben Verresen (205 points) [ revision history ]
edited Feb 28, 2016 by Ruben Verresen

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar\varnothing$sicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...