Yang–Mills and Mass Gap

[arfa brane]

Some cool stuff about linear algebra:

http://jakobschwichtenberg.com/adjoint-representation/
e.g.

It’s often a good idea to look at the Lie algebra of a group to study its properties, because working with a vector space, like $$ TeG $$, is in general easier than working with some curved space, like $$ G $$. An important theorem alled Ado’s Theorem, tells us that every Lie algebra is isomorphic to a matrix Lie algebra. This tells us that the knowledge of ordinary linear algebra is enough to study Lie algebras because every Lie algebra can be viewed as a set of matrices.

A natural idea is now to have a look at the representation of the group $$ G $$ on the only distinguished vector space that comes automatically with each Lie group: The representation on its own tangent vector space at the identity $$ TeG $$, i.e. the Lie algebra of the group!

In other words, in principle, we can look at representations of a given group on any vector space. But there is exactly one distinguished vector space that comes automatically with each group: Its own Lie algebra. This representation is the adjoint representation.

Aha! So that's what is special about an adjoint representation!

 

[QuarkHead]

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?

 

[arfa brane]

In this pdf: http://www.physics.princeton.edu/~mcdonald/examples/EP/mckenzie_lie.pdf, we have the following

A Lie group is a group whose group elements are specified by one or more continuous
parameters which vary smoothly. We consider a simple example, the SO(2) group – rotation in
two dimensions. The group is characterized by a single parameter, θ , the angle of rotation.

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).

 

[hansda]

No, I didn't observe that, because that's wrong. Wavelength is a length (the hint is in the name), not a speed.


This is wrong for similar reasons. Frequency has units of inverse time, not time.


Exactly.


Do you mean: travel a distance of $$cT$$, or travel at the speed $$c$$? And I think both are false conclusions: nowhere did I specify the group velocity of the phase velocity of the wave.


Please explain why.


No, you are making the claim. Please give these equations and calculations yourself. In fact, I don't even know what specific equations you are referring to.

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

 

[NotEinstein]

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

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?

 

[hansda]

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?

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

 

[NotEinstein]

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

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

Edit:

This is wrong for similar reasons. Frequency has units of inverse time, not time.

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?​

 

[NotEinstein]

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.​

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?

 

[QuarkHead]

So there's a homomorphism from U(1) to SO(2)?

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!!

 

[arfa brane]

So try to prove it!

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.

 

[hansda]

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?​

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 $$f=\frac{1}{1}=1 $$.

 

[NotEinstein]

Good. You admitted your error.

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

So now consider $$ fT=1 $$ or $$ f=\frac{1}{T}$$. Here max value of $$ T$$ will be $$T=1 $$.

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

Correspondingly minimum value of $$f $$ will be $$f=\frac{1}{1}=1 $$.

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

 

[hansda]

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

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

 

[NotEinstein]

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

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.

 

[arfa brane]

What Wikipedia says:

The set of all 1×1 unitary matrices clearly coincides with the circle group; the unitary condition is equivalent to the condition that its element have absolute value 1. Therefore, the circle group is canonically isomorphic to U(1), the first unitary group.

The exponential function gives rise to a group homomorphism exp : R → T from the additive real numbers R to the circle group T via the map

The last equality is Euler's formula or the complex exponential. The real number θ corresponds to the angle (in radians) on the unit circle as measured counterclockwise from the positive x-axis. That this map is a homomorphism follows from the fact that the multiplication of unit complex numbers corresponds to addition of angles:

This exponential map is clearly a surjective function from R to T. It is not, however, injective. The kernel of this map is the set of all integer multiples of 2π. By the first isomorphism theorem we then have that

After rescaling we can also say that T is isomorphic to R/Z.

If complex numbers are realized as 2×2 real matrices (see complex number), the unit complex numbers correspond to 2×2 orthogonal matrices with unit determinant. Specifically, we have

The circle group is therefore isomorphic to the special orthogonal group SO(2). This has the geometric interpretation that multiplication by a unit complex number is a proper rotation in the complex plane, and every such rotation is of this form.

--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 . . .

 

[QuarkHead]

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

 

[Farsight]

Interesting stuff.

 

[arfa brane]

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:

4.1 Calculating the Generators

Finding the generators of the Lie algebra from a representation of the Lie group is the first step.
We begin by considering only group elements in the vicinity of the identity element, M(0). We
can expand M(θ) as a Taylor expansion around θ = 0.

Pedantically, the expansion is a MacLaurin series, in that case?

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

 

[arfa brane]

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.

 

[arfa brane]

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)

​

In which case, the complex number (as a pair of real numbers), is at the intersection of A and B (the point where all four corners of the rectangle meet).
But again, it's the same result as the first method, up to a scaling factor.
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).