After a comment of John Baez to a question I asked on MathOverflow, I would like to ask what the difference between, for example, $SU(3)\times SU(2) \times U(1) $ and $SU(3) \otimes SU(2) \otimes U(1)$ is. The $\times$ is the Cartesian product while the $\otimes$ is the tensor product. I gave the example of the Standard Model gauge group but it can be any product of groups. My question is when talking about global or gauge groups, do we mean Cartesian products or tensor products? And what is the real difference between them anyway?
Group Theory – Difference Between Cartesian Product and Tensor Product on Groups
gauge-theorygroup-theorynotationrepresentation-theorytensor-calculus
Related Solutions
Baez actually has another paper (with Huerta) that goes into more detail about this. In particular, Sec. 3.1 is where it's explained, along with some nice examples. The upshot is that the hypercharges of known particles work out just right so that the action of that generator is trivial. Specifically, we have
Left-handed quark Y = even integer + 1/3
Left-handed lepton Y = odd integer
Right-handed quark Y = odd integer + 1/3
Right-handed lepton Y = even integer
Because these are the only values for known fermions in the Standard Model, that generator does nothing. So basically, you can just take the full group modulo the subgroup generated by $(\alpha, \alpha^{-3}, \alpha^2)$ -- where $\alpha$ is a sixth root of unity.
There's also this paper by Saller, which goes into greater detail about the "central correlations" of the Standard Model's gauge group, but in a more technical presentation. Saller also goes into some detail in chapter 6.5.3 of his book.
$\lvert A\rangle \langle B \rvert$ is the tensor of a ket and a bra (well, duh). This means it is an element of the tensor product of a Hilbert space $H_1$ (that's where the kets live) and of a dual of a Hilbert space $H_2^\ast$, which is where the bras live. Although for Hilbert spaces their duals are isomorphic to the original space, this distinction should be kept in mind. So we can "feed" a ket $\lvert \psi\rangle$ from $H_2$ to the bra in $\lvert \phi\rangle\otimes \langle\chi\rvert \in H_1\otimes H_2^\ast$, and are left with a state in $H_1$ given by $\langle \chi \vert \psi\rangle \lvert \phi\rangle$. The usual use case for such a tensor product is when $H_1=H_2$ to construct a map from $H_1$ to itself, e.g. the projector onto a state $\lvert \psi \rangle$ is given by $\lvert\psi\rangle \langle \psi \rvert$.
In general, a tensor in $H_2 \otimes H_1^\ast$ corresponds to a linear operator $H_1\to H_2$. In the finite-dimensional case, these are all linear operators, in the infinte-dimensional case, this is no longer true, e.g. $H^\ast \otimes H$ are precisely the Hilbert-Schmidt operators on $H$.
In constract, a tensor $\lvert A\rangle\otimes \lvert B\rangle$ (also just written $\lvert A \rangle \lvert B\rangle$) in $H_1\otimes H_2$, although it corresponds to a bilinear map $H_1\times H_2\to\mathbb{C}$ by definition, is usually not meant to denote an operator, but a state. Given two quantum systems $H_1$ and $H_2$, $H_1\otimes H_2$ is the space of the states of the combined system (as for why, see this question).
Best Answer
I) The main point is that we usually only consider tensor products $V \otimes W$ of vector spaces $V$, $W$ (as opposed to general sets $V$, $W$). But groups (say $G$, $H$) are often not vector spaces. If we only consider tensor products of vector spaces, then the object $G \otimes H$ is nonsense, mathematically speaking.
With further assumptions on the groups $G$ and $H$, it is sometimes possible to define a tensor product $G \otimes H$ of groups, cf. my Phys.SE answer here and links therein.
II) If $V$ and $W$ are two vector spaces, then the tensor product $V \otimes W$ is again a vector space. Also the direct or Cartesian product $V\times W$ of vector spaces is isomorphic to the direct sum $V \oplus W$ of vector spaces, which is again a vector space.
In fact, if $V$ is a representation space for the group $G$, and $W$ is a representation space for the group $H$, then both the tensor product $V\otimes W$ and the direct sum $V\oplus W$ are representation spaces for the Cartesian product group $G\times H$.
(The direct sum representation space $V\oplus W\cong (V\otimes \mathbb{F}) \oplus(\mathbb{F}\otimes W)$ for the Cartesian product group $G\times H$ can be viewed as a direct sum of two $G\times H$ representation spaces, and is hence a composite concept. Recall that any group has a trivial representation.)
This interplay between the tensor product $V\otimes W$ and the Cartesian product $G\times H$ may persuade some authors into using the misleading notation $G\otimes H$ for the Cartesian product $G\times H$. Unfortunately, this often happens in physics and in category theory.
III) In contrast to groups, note that Lie algebras (say $\mathfrak{g}$, $\mathfrak{h}$) are always vector spaces, so tensor products $\mathfrak{g}\otimes\mathfrak{h}$ of Lie algebras do make sense. However due to exponentiation, it is typically the direct sum $\mathfrak{g}\oplus\mathfrak{h}$ of Lie algebras that is relevant. If $\exp:\mathfrak{g}\to G$ and $\exp:\mathfrak{h}\to H$ denote exponential maps, then $\exp:\mathfrak{g}\oplus\mathfrak{h}\to G\times H$.