# Dimension of tensor product vs. dimension of direct product

I am really sorry if this is trivial but I come from Physics and it is really confusing for me to understand what is going on and looking at the answers on both Physics and Math SE sent me through a rabbit hole and I am more confused than ever. From what I understand from this this MathSE answer is that for finite sequence of vector spaces the direct product is essentially the same as the direct sum with the added property of linearity.

A tensor product, however, is entirely different. Most Physics texts tend to use both $$\otimes$$ and $$\times$$ to denote the tensor(?) product. So if when I say something like $$SU(2)\times SU(2)$$, is it actually $$SU(2)\otimes SU(2)$$? So, roughly, the statement of addition of angular momenta is that a tensor product state can be decomposed as a direct sum.

Next, what made me even more confused is that, I saw it in Matthew D. Schwartz's QFT and the Standard Model that the Lie Algebra of the Lorentz group (which is $$SU(2)\otimes(?)SU(2)$$) breaks down into $$\mathfrak{so}(1,3) = \mathfrak{su}(2)\oplus\mathfrak{su}(2)$$ where $$\mathfrak{su}(2)$$ is the Lie Algebra of $$SU(2)$$. But that would mean that if I have matrix respresentations of Lie Algebras $$\mathfrak{h},\mathfrak{g}$$ with dimensions $$N$$ and $$M$$ respectively, their direct sum has dimension $$N+M$$ and so the representations of the group obtained from this representation of the algebra also has dimension $$N+M$$ so are we talking about the direct product here? So the $$SO^{+}(1,3) \cong SU(2)\times SU(2)$$, actually? I am so confused.

I got downvoted in PhysicsSE, can someone please explain to me what is the distinction and if the above is true for Lie Algebras? I am absolutely clueless about this, thanks.

• I would expect $A \times B$ to the be the cartesian product of and $A \otimes B$ to be the tensor product. Their algebraic structure are significantly different. Feb 9 at 21:34
• Every bilinear function on $V\times W$ extends to a unique linear function on $V\otimes W$. Feb 9 at 21:34
• "Most Physics texts tend to use both $\otimes$ and $\times$ to denote the tensor product" - I strongly doubt this. Feb 9 at 21:51
• @Filippo see for example, Wu Ki Tung's Group Theory in Physics. Feb 9 at 22:01
• Tensor products are for vector spaces or algebras. $SU(2)$ is a Lie group. There is no recognised definition for the tensor product of Lie groups. According to math.stackexchange.com/questions/2880510/…, there are circumstances in which the tensor product (qua algebras) of two Lie algebras can be made into a Lie algebra. Feb 9 at 22:01

It seems to me that one thing you’re confusing is groups and their representations. There’s no such thing as a tensor product of groups, so $$SU(2)\otimes SU(2)$$ makes no sense. There are various products of groups, but the one that’s usually meant by ”product of groups”, and that’s denoted by $$\times$$, is the direct product of groups.

Given representations $$U$$ and $$V$$ of groups $$G$$ and $$H$$ (i.e. vector spaces equipped with linear group actions), both the tensor product $$U\otimes V$$ and the direct sum $$U\oplus V$$ naturally carry a representation of the direct product $$G\times H$$: $$(g,h)(u\otimes v)=(gu)\otimes(hv)$$ and $$(g,h)(u\oplus v)=(gu)\oplus (hv)$$. Usually, the tensor product representation is more useful; this is explained here: direct sum of representation of product groups.

Especially in physics, where taking the tensor product of Hilbert spaces corresponds to combining systems, it’s usually the tensor product of representations that’s of interest. The dimension of the representation and the dimension of the Lie algebra are two different things. There’s no contradiction between forming the direct sum of Lie algebras, whose dimension as a vector space is the sum of the dimensions of the summands as vector spaces, and the tensor product of their representations, whose dimension is the product of the dimensions of the factors.

Then there’s the connection between the direct product of Lie groups and the direct sum of their Lie algebras: $\operatorname{Lie}(G \times H)\cong \operatorname{Lie}(G)\oplus \operatorname{Lie}(H)$. This is perhaps confusing in that the operation on the left is typically referred to as a product and the one on the right is typically referred to as a sum. As you say, for finitely many factors/summands, these are the same. I think the reason that “direct product” is preferred for groups whereas “direct sum” is preferred for vector spaces is that the direct product is the categorical product in the category of groups, and it’s not the coproduct (sometimes also called the categorical sum), which is, rather, the free product. By contrast, in the category of vector spaces the direct sum is the coproduct (or categorical sum).

• Sorry, in this context I was viewing $SU(2)$ as a matrix group. Feb 9 at 22:25
• @QFTheorist: I don’t understand. $SU(2)$ is a matrix group. What part of what I wrote did you take to contradict that? Feb 9 at 22:29
• @QFTheorist: Ah, I think I see what caused the misunderstanding. A representation is a vector space together with a linear group action on that vector space. As described in Chris’ answer, a linear group action can be thought of as a homomorphism from the group to a matrix group. But the tensor product of representations is the tensor product of the underlying vector spaces (equipped with the linear group action given in the answer), not of the linear groups or matrix groups. (And when I talk about the dimension of a representation, I mean the dimension of the underlying vector space.) Feb 9 at 22:38
• Thanks @joriki . That was exactly what I was confused about. Feb 9 at 22:40

Well, first, as Lie groups, it does not in full generality really make sense to take tensor products. Given a two Lie groups, we take their cartesian product (also known as direct product) which is again Lie group. If $$G$$ and $$H$$ are Lie groups, $$G\times H$$ is a smooth manifold with the product manifold structure, and a group with the product group structure that is compatible with the product manifold structure so it is indeed a Lie group.

Now, let $$V$$ be a vector space. Then a $$\textbf{representation}$$ of a Lie group $$G$$ on $$V$$, is a smooth homomorphism $$\rho:G\rightarrow GL(V)$$. Ok, let's unpack that jargon for a second. The object $$GL(V)$$ is just the general linear group of $$V$$, that is the group of linear maps $$V\rightarrow V$$ which are also invertible. If $$V$$ is a finite dimensional vector space, then after chosing a basis $$\{e_i\}$$, this just become the general linear group $$GL(\mathbb R^n)$$, or $$GL(\mathbb C^n)$$ depending on whether $$V$$ is a real or complex vector space. These are precisely the groups of invertible $$n\times n$$ matrices over $$\mathbb R$$ and $$\mathbb C$$ respectively. For $$\rho$$ to be smooth mean's just that, the map is differentiable. Of $$\rho$$ to be a homomorphism, means that $$\rho(g\cdot g')=\rho(g)\cdot \rho (g')$$, so $$\rho$$ respects the group structures of $$G$$ and $$GL(V)$$.

As an example, the matrix Lie groups such as $$GL(\mathbb R^n)$$, $$O(n)$$, $$SO(t,s)$$, and $$U(n)$$ all have obvious representations on the vector spaces $$\mathbb R^n$$, and $$\mathbb C$$ (where $$s+t=n$$), because they are precisely defined to be subsets of the vector spaces of $$n\times n$$ matrices.

When we have a representation of $$G$$ on $$V$$, this means that we can take an element $$g\in G$$, and element in $$v\in V$$, and a get a new element $$\rho(g)\cdot v\in V$$, which we often just denote by $$g\cdot v$$.

For finite products, the direct product, and the direct sum are the same, but the tensor product is wholly different. With out getting to universal properties and what not (though I suggest you look into that as it makes tensor products, direct sums, and direct products make way more sense) $$V\times W$$, or $$V\oplus W$$ are essentially just pairs $$(v,w)$$, for $$v\in V$$, and $$w\in W$$, where vector additions is defined by $$(v,w)+(v',w')=(v+v',w+w')$$. A tensor product, is however only $$\textit{generated}$$ by such pairs. That means, we have the pairs $$v\otimes w$$, and $$v'\otimes w'$$, but there is, without knowing more about what $$v,v',w$$ and $$w'$$ are, not general simplication for the expression $$v\otimes w+v'\otimes w'$$.

If $$V$$ and $$W$$ have basises $$\{e_i: 0\leq i\leq n\}$$ and $$\{f_j:0\leq j\leq m\}$$ for some $$n$$ and $$m$$ in $$\mathbb N$$, then a basis for $$V\oplus W$$ is the set $$\{(e_i,0), (0,f_j): 0\leq i\leq n, 0\leq j\leq m\}$$, while a basis for $$V\otimes W$$ is the set $$\{e_i\otimes f_j: 0\leq i\leq n, 0\leq j\leq m\}$$. In the direct sum, we clearly have $$m+n$$ basis vectors, but in the tensor product, we have $$n\cdot m$$ basis vectors. So direct sums create vector spaces of dimension $$\dim V+\dim W$$, while tensor products create vector spaces of dimension $$\dim V\cdot \dim W$$.

If we have representations of $$G$$ on $$V$$ and $$W$$, then we have induced representations on $$V\oplus W$$ and $$V\otimes W$$ given by $$g\cdot (v,w)=(g\cdot v,g\cdot w)$$, and $$g\cdot (v\otimes w)=(g\cdot v)\otimes (g\cdot w)$$.

Now that we have gotten the representation theory out of the way, it is time to address the actual statements (which are often times completely incorrect) that physicists make about Lie groups, and their Lie algebras. To every Lie group, there is a Lie algebra, which is a vector space with with a commutator bracket that is isomorphic to the tangent space at the identity of a Lie group. You can explicitly calculate the Lie algebras of your favorite Lie groups by taking their defining properties, differentiating at the identity, and then applying a dimension argument. For example, if $$\gamma$$ is a curve in $$O(n)$$ passing through the identity, such that its tangent vector at $$\gamma(0)=I$$ is $$X$$, then we have that: $$\gamma(t)\cdot \gamma(t)^T=I$$ Taking a time derivative at $$t=0$$, we have that: $$X+X^T=0$$ A dimension argument then demonstrates that the Lie algebra of $$O(n)$$ is the vector subspaces of antisymmetric $$n\times n$$ matrices.

So, first and foremost, no one means $$SU(2)\otimes SU(2)$$, I personally don't even know what such a statement would mean, unless we're using the group to refer to a tensor product representation, which I think is bad notational practice. What one would mean is $$SU(2)\times SU(2)$$.

Secondly, the statement: $$\mathfrak{so}(1,3)\cong \mathfrak{su}(2)\oplus \mathfrak{su}(2)$$ is completely false. It is also not true that $$SO(1,3)\cong SU(2)\times SU(2)$$, and even if the statement of Lie algebras was true, this would not imply the statement about the groups, as non isomorphic Lie groups have isomorphic Lie algebras as we are about to see.

What is true, is that the Lie algebra of $$SO(t,s)$$ is isomorphic to the Lie algebra of the spin group $$\operatorname{Spin}(t,s)$$ for all signatures of pseudo euclidean inner products. The spin group is the double cover of $$SO^+(t,s)$$ (that is the special orthogonal matrices which are also time orientable), and what we mean by that is that there is a $$2$$ to $$1$$ group homomorphism $$\operatorname{Spin}(t,s)\rightarrow SO(t,s)$$.

In special cases, we have "exceptional isomorphisms", so for example the spin group of $$SO(3)$$ is $$SU(2)$$, so the lie algebras $$\mathfrak{so}(3)$$ and $$\mathfrak{su}(2)$$ are isomrophic. The spin group of $$SO^+(1,3)$$ is $$SL(2,\mathbb C)$$ so what is true is that $$\mathfrak{so}(1,3)\cong \mathfrak{sl}(2,\mathbb C)$$. Moreover, confusingly enough in your case, the spin group of $$SO(4)$$ is $$SU(2)\times SU(2)$$, so the Lie algebras $$\mathfrak{so}(4)$$ is isomorphic to $$\mathfrak{su}(2)\oplus \mathfrak{su}(2)$$. Note that none of these groups are isomorphic to one another, but the Lie algebras are.

As a quick aside, I will sketch the $$2$$ to $$1$$ map $$SL(2,\mathbb C)\rightarrow SO^+(1,3)$$ for you. Identify $$\mathbb R^{1,3}$$ with a real subspace of $$2\times 2$$ complex matrices given by the map: $$v=(t,x,y,z)\longmapsto X_v=\begin{pmatrix} t+z&x-iy\\ x+iy&t-z \end{pmatrix}$$ Then $$\det X=t^2-x^2-y^2-z^2=\eta(v,v)$$, so we define a map: \begin{align} \pi:SL(2,\mathbb C)\longrightarrow SO^+(1,3)\\ A&\longmapsto \pi(A) \end{align} where $$\pi(A)$$ acts $$v\in \mathbb R^{1,3}$$ via: $$\pi(A)\cdot v= AX_vA^\dagger$$ This clearly preserves the norm on $$\mathbb R^{1,3}$$, so $$\pi(A)$$ really is an element in $$O(1,3)$$, checking that it is orientable and time orientable I leave as an exercise to you, as the computations get a bit hairy. The other $$2$$ to $$1$$ maps are defined similarly.

To end the long answer, my guess is that the author is incorrectly stating that $$\mathfrak{su}(2)\oplus \mathfrak{su}(2)\cong \mathfrak{so}(1,3)$$, because many problems in QFT can be solved in Euclidean signature via a wick rotation, so maybe intuitively there is a nice identification going on? I am no expert on QFT, so I can't be sure, but I know for a fact that that statement about Lie algebras as currently formulated is incorrect.