Question about the Lie group $SU(3) \times SU(2) \times U(1)$ and the concept of manifold

Question

I don't know if this question is a duplicate, so I'll delete if is.

Well, I'm in the very beginning of the study of contemporary topics such as gauge theories, I would say that I'm in a "science communication + General Relativity mathematical/physical background" level.

So, in the book $[1]$, in chapter 1 introduction, the author wrote:

From a mathematical point of view, continuous symmetry groups can be conceptualised as Lie groups. By definition, Lie groups are groups in an algebraic sense which are at the same time smooth manifolds, so that both structures – algebraic and differentiable – are compatible. $\tag{1}$

and then he stated:

The currently accepted Standard Model of elementary particles, for instance, is a gauge theory with Lie group:

$$ SU(3) \times SU(2) \times U(1) \tag{2} $$

Now, in GR, the spacetime is a manifold and the points of spacetime are events. Intuitively, an event is a 4-dimensional "point" where a particle can be at.

Well, based on $(1)$, we can say that Lie groups are manifolds; based on $(2)$, we can say that $SU(3) \times SU(2) \times U(1)$ is a Lie group and therefore a manifold.

So, what is supposed to mean (physically) a point on the manifold $SU(3) \times SU(2) \times U(1)$?

$$ * * * $$

$[1]$ HAMILTON. J.D.M, Mathematical Gauge Theory With Applications to the Standard Model of Particle Physics, Springer, 2017.

Answer 1

Let me begin by answering the simpler question "What is $U(1)$ as a manifold?" Recall that the elements of $U(1)$ are $1\times1$, complex, unitary matrices. If $g$ is a generic $1\times1$, complex matrix, then we can write \begin{align} g = \begin{pmatrix} r e^{i\varphi} \end{pmatrix} \end{align} for real numbers $r$, $\varphi$. A unitary matrix obeys $U^{\dagger}U = 1$, so a $1\times1$ unitary matrix $g$ will have $r=1$. That is \begin{align} g = \begin{pmatrix} e^{i\varphi} \end{pmatrix} \end{align} We see that we can parameterize $U(1)$ by a single real number $\varphi$. Since $g$ only depends on $e^{i\varphi}$, the parameters $\varphi$ and $\varphi + 2\pi$ label the same group element. So we can say that $U(1)$ as a manifold is the same as the space of real numbers $\varphi$ with the identification $\varphi \simeq \varphi + 2\pi$. This space is precisely the circle $\mathbb{T} \simeq S^1$.

We can do basically the same thing to find a parametrization for $SU(2)$. A generic element $g$ of $SU(2)$ can be written in terms of four real numbers $x$, $y$, $z$, $w$ as \begin{align} g = \begin{pmatrix} x + iy & -z + iw\\ z+iw & x - iy \end{pmatrix}, \end{align} where $x^2 + y^2 + z^2 + w^2 = 1$. This latter constraint comes from demanding that $\det g = 1$. The space of points $(x,y,z,w) \in \mathbb{R}^4$ obeying $x^2 + y^2 + z^2 + w^2 = 1$ is the 3-sphere $S^3$. So as a manifold we identify $SU(2)$ with $S^3$.

$SU(3)$ is topologically more complicated than the lower-dimensional cases. Suffice it to say that we can label points on $SU(3)$ by a set of real numbers with some relations between them, as with $U(1)$ and $SU(2)$.

For two groups $G$ and $H$, the direct product $G\times H$ consists of pairs $(g,h)$ of elements $g\in G$, $h\in H$. Similarly the direct product $G_1 \times G_2 \times G_3 \times \ldots$ consists of tuples $(g_1, g_2, g_3, \ldots)$.

So an element of $U(1) \times SU(2) \times SU(3)$ is a tuple $(g_1, g_2, g_3)$ where $g_1 \in U(1)$ is a point on the circle, $g_2 \in SU(2)$ a point on the 3-sphere, and $g_3 \in SU(3)$ a point in $SU(3)$.