The h Bar

unitary matrices have $|\det(U)| = 1$

So $\det(U) = e^{i\theta}$

so $U(3)$ and $SU(3)$ are isomorphic?

Well no, not quite

as an example

$U(1)$ is the circle

$SU(1)$ is $I$

hmm

is there another property that $SU(1)$ has that $U(1)$ doesn't that distinguishes them? because I was under the impression that it was just the unitary determinant

but apparently they both have that property

There might be something like $U(N) \approx S \times SU(N)$?

1:27 AM

is that the homeomorphism symbol?

since for any element of $U(N)$ on a circle there corresponds one on $SU(N)$

No that's just me going "Eeeeh, maybe?"

ah ok

Lemme check the book of lie stuff

Unitary group

In mathematics, the unitary group of degree n, denoted U(n), is the group of n × n unitary matrices, with the group operation of matrix multiplication. The unitary group is a subgroup of the general linear group GL(n, C). Hyperorthogonal group is an archaic name for the unitary group, especially over finite fields. For the group of unitary matrices with determinant 1, see Special unitary group. In the simple case n = 1, the group U(1) corresponds to the circle group, consisting of all complex numbers with absolute value 1 under multiplication. All the unitary groups contain copies of this group...

ah yes, there is a connection

in this context do you mean connection as in there is a link or connection as in the connection on a manifold

I haven't really dug into lie groups yet, this was just something I saw offhandedly that surprised me

Q: Quotient of unitary group by special unitary group

$SU(N)$ is a subgroup of $U(N)$

so it's not terribly surprising that they both obey that property

Semiclassical

1:36 AM

U(N)/SU(N) ~ S, maybe?

what is S?

circle group

ah

ah, yes:

1

Can someone help me, I don't understand the following question. "Using without proof, the homomorphism theorem, or otherwise, show that $U(n)/SU(n)$ is isomorphic to $U(1)$." Here, $U(n)$ is the unitary group while $SU(n)$ is the special unitary group.

group-theory

$S$ is, as the letter indicates, the sircle

4

2:03 AM

So I have trained a neural network on TeX formulas

It's a bit early in the training, but so far, here's a sample :

\begin{eqnarray}
m (\lambda_p, \lambda_0) &=& \lambda_0 \ldots &=& \mu_g(\lambda_p) \ldots\\
&=& \lambda_g(\lambda_p)
\end{eqnarray}
\begin{eqnarray}
m (\lambda_0, \lambda_g) &=& \lambda_g(\lambda_0) \ldots &=& \lambda_g(\lambda_0)
\end{eqnarray}
\begin{eqnarray}
\sigma^{-1}(\lambda_p, \lambda_0) &=& \lambda_0 \ldots &=& \sigma^{-1}(\lambda_p, \lambda_0)
\end{eqnarray}

Mysterious formulas

@Charlie best seen using U(2) and SU(2)..

Take SU(2)... you can write such a matrix easily in the standard parametrization...

multiply all entries by some constant phase factor...

\begin{eqnarray}
[\mu_{\mu\nu} (\varepsilon + \frac{1}{2}(a^a (x_2, e_1))] &=& |(\kappa^a|^2) \sigma^a (x_2, e_1)\\
\end{eqnarray}
\begin{eqnarray}
[\mu_{\mu\nu} (\varepsilon + \frac{1}{2}(\kappa^a (x_2, e_1))] &=& \varepsilon \sigma^a (x_2, e_1)\\
\end{eqnarray}
\begin{eqnarray}
[\mu_{\mu\nu} (\varepsilon + \frac{1}{2}(\kappa^a (x_2, e_1))] &=& \varepsilon + |\sigma^{a}|^2 (x_2, e_1)\\
\end{eqnarray}
\begin{eqnarray}
[\mu_{\mu\nu} (\varepsilon, \frac{1}{2}(\kappa^a (x_2, e_1))] &=& \varepsilon + |\sigma^{a}|^2 e_1 (- \varepsilon + \frac{1}{2}(\kappa^a (x_2, e_1))\\

the unitarity property remains $M^\dagger M=$unit...

Not bad for just a hundred cycles!

but now det(M) is multiplied by square of that phase factor.

2:09 AM

but the det(M) is still 1 right?

oh wait