# Yang–Mills and Mass Gap

Discussion in 'Physics & Math' started by Thales, Nov 29, 2017.

1. ### arfa branecall me arfValued Senior Member

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Some cool stuff about linear algebra:

e.g.

Last edited: Feb 14, 2018

3. ### QuarkHeadRemedial Math StudentValued Senior Member

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Yes. To complete the picture.......

Every Lie group is manifold (it's the definition). The realization of a group is the set of all mappings from the manifold to itself.

The representation of a group is the set of all mappings from a vector space to itself (this is usually the embedding space).

Since a Lie group is also a manifold, it comes equipped with a vector space tangent to every poInt. The adjoint representation is the set of all such mappings where the vector space is tangent at the group identity.

This is called the Lie algebra associated to the group

Here's what is remarkable about the Standard Model - every vector space must have a set of basis vectors. The number of such basis vectors for the vector space at the identity determines the number of gauge bosons for each sector - 1 for EM (U(1)), 3 for the weak nuclear (SU(2)) and 8 for the strong nuclear (SU(3)).

How cool is that?

5. ### arfa branecall me arfValued Senior Member

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In this pdf: http://www.physics.princeton.edu/~mcdonald/examples/EP/mckenzie_lie.pdf, we have the following
So there's a homomorphism from U(1) to SO(2)? Both are Lie groups because the parameter, θ, is continuous, hence we can consider θ + dθ, or we can make θ as small as we like, or "close to the identity"? Of course that means the identity element is some choice where we set θ = 0?

The manifold is just the unit circle (I guess that's why U(1) is known as the circle group).

7. ### hansdaValued Senior Member

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You can check with the wave equations here https://en.wikipedia.org/wiki/Wave . Particularly see the "Sinosoidal Waves" section.

8. ### NotEinsteinValued Senior Member

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I'm already fully familiar with that, yes, and I suggest you get so too. But how is this a response to my post? Can you actually answer any of my questions? Can you actually justify any of your claims?

9. ### hansdaValued Senior Member

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If you see the "Sinosoidal Section" there, you will observe that $fT=1$. But you consider this as wrong.

10. ### NotEinsteinValued Senior Member

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Can you please post a quote where I have said this, because I don't remember saying that?

Edit:
Ah, there it is. Yes, that is a mistake. (I don't know what happened; I guess I saw a dividing-slash or something?)

Obviously I'm wrong there, as I myself immediately confirm in the next sentence in that post.

Thank you for pointing out this mistake. Now can you reply to my post?

Last edited: Feb 18, 2018
11. ### NotEinsteinValued Senior Member

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I've went through my old notes, and I see what happened. You initially wrote "f = T", to which I responded as I did. You must have corrected that mistake within a few minutes (no "edit"-timestamp displayed), so when I went to type up my reply I quoted the corrected version. So it's more a typo you made than I mistake I made. Anyway, now can you reply to my post?

12. ### QuarkHeadRemedial Math StudentValued Senior Member

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So try to prove it!

Recall that $U (1)$ has $e^{i\theta}$ as a group element for any Real $\theta$

Recall also that an element in $SO(2)$ is $\begin {pmatrix}\cos \theta&&\sin \theta\\-\sin\theta &&cos \theta \end{pmatrix}$

Finally recall that $e^{i \theta}= \cos \theta+ i\sin \theta$

Go!!

13. ### arfa branecall me arfValued Senior Member

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Ok, first up, the element of SO(2) you've written I would say is the inverse of an element of SO(2). But that's just a choice of whether an element rotates a point in a clockwise or counter-clockwise direction.

Since a complex number can be represented by a 2 x 2 matrix, we should have $e^{i \theta} = cos \theta + i sin \theta$ has a matrix representation as $M(\theta) = \begin{pmatrix} cos \theta & -sin \theta \\ sin \theta & cos \theta \end{pmatrix}$.

So $e^{i \theta} e^{i \phi} = e^{i (\theta + \phi)} = cos (\theta + \phi) + i sin (\theta + \phi) = \begin{pmatrix} cos (\theta + \phi) & - sin (\theta + \phi) \\ sin (\theta + \phi) & cos (\theta + \phi) \end{pmatrix}$.

But since $cos (\theta + \phi) = cos \theta cos \phi - sin \theta sin \phi$,
and $sin (\theta + \phi) = sin \theta cos \phi + cos \theta sin \phi$,

we have: $\begin{pmatrix} cos (\theta + \phi) & - sin (\theta + \phi) \\ sin (\theta + \phi) & cos (\theta + \phi) \end{pmatrix} = \begin{pmatrix} cos \theta & -sin \theta \\ sin \theta & cos \theta \end{pmatrix} \begin{pmatrix} cos \phi & -sin \phi \\ sin \phi & cos \phi \end{pmatrix} = M(\theta)M(\phi)$.

And clearly $\theta = \phi = 0$ takes the identity in U(1) to the identity in SO(2).

I think that's pretty much it.

Last edited: Feb 19, 2018
14. ### hansdaValued Senior Member

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Good. You admitted your error. So now consider $fT=1$ or $f=\frac{1}{T}$. Here max value of $T$ will be $T=1$. Correspondingly minimum value of $f$ will be $f=\frac{1}{1}=1$.

15. ### NotEinsteinValued Senior Member

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As I explained, I'm not sure it was mine, but OK.

This is wrong. $T$ can be higher than $1$.

Sure, for a maximum value of $1$ for $T$, that's correct. But $1$ isn't the maximum value of $T$.

Last edited: Feb 19, 2018
16. ### hansdaValued Senior Member

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Can you explain, how $T$ can be higher than $1$?

17. ### NotEinsteinValued Senior Member

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I've already given at least one example earlier, but sure, here's another one: let's say a sphere rotates once per hour. 1 hours is 60 minutes, so $T=60$, which is larger than $1$. QED.

18. ### arfa branecall me arfValued Senior Member

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What Wikipedia says:
--https://en.wikipedia.org/wiki/Circle_group

. . . and the reason the exponential map from R to T is not injective (1 to 1), is that there is more than one sum of angles with the same value. The reason the Lie algebra is trivial is that there is only one parameter . . .

Last edited: Feb 19, 2018
19. ### QuarkHeadRemedial Math StudentValued Senior Member

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It is better than that (or worse, according to your perspective).

Suppose Lie algebra $\mathfrak{g}$. Then the group $G$ is recovered via the exponential map (a short session of beard tugging should convince this is a "sort of inverse" of a Taylor series).

Very nicely, suppose $g \in G$ and $X \in \mathfrak{g}$. Then there is a Lemma (I don't recall whose) that states that $\det g = e^{tr X}$.

Incidentally, your proof of homomorphism could have been done more economically , I think, but basically looked OK. But didn't prove preservation of inverses. And you merely asserted preservation of indentities.

Of course for abelian groups, given the one the other comes for free

20. ### Farsight

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Interesting stuff.

21. ### arfa branecall me arfValued Senior Member

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Wikipedia says that in fact $e^{i\theta} = cos\theta + isin \theta$ is a homomorphism from U(1) to "the circle group" T, where T is a 1-torus. It's interesting too, that a complex number represents (geometrically) a point on a 2-torus, where we can take, say $\alpha + \beta i$ to be a pair of real angles, one of which is multiplied by i. The angles define distinct points on the pair of circles whose product $S^1 \times S^1$ is the torus.

And the homomorphism in my (not very rigorous) proof, is clearly an isomorphism (thus says Wikipedia, since the mapping $e^{i\theta} \mapsto M(\theta)$ is "canonical").

From http://www.physics.princeton.edu/~mcdonald/examples/EP/mckenzie_lie.pdf:
Pedantically, the expansion is a MacLaurin series, in that case?

Ed: yes, but a Maclaurin series (no capital L!) is a Taylor series, duh.

Last edited: Feb 21, 2018
22. ### arfa branecall me arfValued Senior Member

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Well, I have been tugging my beard (because I actually have one), and the idea of a "sort of" inverse of a Taylor expansion hasn't given me much, except I can perhaps start with a decomposition of an element of SO(2) into a sum of two matrices, because, well, because I can decompose a matrix into a sum of matrices.

Such that: $\begin{pmatrix} cos \theta & -sin \theta \\ sin \theta & cos \theta \end{pmatrix} = \begin{pmatrix} cos \theta & 0 \\ 0 & cos \theta \end{pmatrix} + \begin{pmatrix} 0 & -sin \theta \\ sin \theta & 0\end{pmatrix}$.

Then: $\begin{pmatrix} cos \theta & 0 \\ 0 & cos \theta \end{pmatrix} + \begin{pmatrix} 0 & -sin \theta \\ sin \theta & 0\end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}cos \theta + \begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix} sin \theta$.

So clearly, the identity in SO(2) multiplies the cos term, and a different matrix multiplies the sin term; the latter is in fact the key to the whole "infinitesimal rotation" algebra, since the last eqn can be rewritten thus:

$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}cos \theta + i\begin{pmatrix} 0 & i \\ -i & 0\end{pmatrix} sin \theta$.

Thus, I get an Hermitian matrix: $\begin{pmatrix} 0 & i \\ -i & 0\end{pmatrix}$. Suppose I call this X (because that's what McKenzie calls it).

I'm busy with other stuff, so that's it for now.

Last edited: Feb 23, 2018
23. ### arfa branecall me arfValued Senior Member

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Another aside about how a single complex number can be considered a pair of angles which map to a 2-torus. Clearly that means whatever value the pair of real numbers have which are mapped to angles, you have to make each real number "fit" the interval $[0, \;2\pi]$, you could take the remainder say, after division of both the numbers by $2\pi n$ for n an arbitrary positive integer (not zero).

You could also construct a rectangle with sides equal to the real number values and glue opposite sides together: (do the reverse of what this diagram says)

This scaling is what's known otherwise as gauging, and is one reason why $e^{i\theta}$ is an element of a gauge group of symmetries.
In QM, $e^{i\theta}$ doesn't scale things , but represents a choice of phase for say, the electron matter-field (described by a wavefunction of course).