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 $SU(3)$.

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

I think this is not quite the question you meant to ask. Minkowski space has the 10-dimensional Poincaré group as its symmetry group, and GR arguably has the infinite-dimensional diffeomorphism group as a symmetry group, but spacetime isn't either of those manifolds; it's 4-dimensional. I think what you mean to ask is: what is the manifold of which $SU(3) \times SU(2) \times U(1)$ is the symmetry group, and what is the meaning of points on that manifold.

The idea that the gauge group is literally the symmetry group of extra spatial dimensions is called Kaluza-Klein theory, and it's taken very seriously in quantum gravity, although there's no direct evidence for it and no hint of which of the many manifolds with the appropriate symmetries might be the correct one. One especially simple possibility comes from SO(10) grand unification, where the Standard Model gauge group is a subgroup of the continuous isometries of a 9-sphere (or Spin(10), its double cover), suggesting 9 extra dimensions with a uniform positive curvature. The minimum possible dimensionality of a manifold with the necessary symmetries is 7, which is interesting because $4+7=11$ is an important number in string theory.

In principle the extra dimensions are like the big three dimensions and the points are just points. However, all known particles are spread around the extra dimensions much like low-energy orbitals of the hydrogen atom, and therefore useless for probing the shape. The curvature radius of the extra dimensions may be comparable to the Planck length anyway, and they may not actually resemble space as we know it.