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

Last edited: Feb 14, 2018 at 7:56 AM

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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?

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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).

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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. ### NotEinsteinRegistered 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. ### NotEinsteinRegistered 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 at 4:53 PM