Tensor #20. Algebraic Structures II: Modules to Algebras텐서 #20. 대수 구조 II: 모듈에서 대수까지

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 20 of the playlist, Algebraic Structures II - Modules to Algebras. The previous post climbed through the structures with one set, and this time we move to structures with two sets and end by building the tensor algebra.

Why we need two sets

The structures of the previous post split into a group-like family with one operation and a ring-like family with two, but all of them have just one set. The operations were only of the form $G \times G \to G$. A vector space is different. Its vectors form an abelian group under addition, but scalar multiplication $\mathbb{R} \times V \to V$ brings in another set, the real numbers. In a structure with only one set such a map cannot even be defined, so a vector space is neither group-like nor ring-like but a more complicated object.

Module

A structure more general than a vector space is a module. It has two ingredients. One is an abelian group $G$, and the other is a ring $R$ with addition and multiplication. On top of them we add a map that takes an element of the ring and an element of the group and returns an element of the group.

If the ring multiplies from the left this is a left $R$-module, and if the product has the form $G \times R \to G$ it is a right module. The distinction arises because the ring need not be commutative. This map must obey the following linearity conditions to make a module.

These three equations mix different additions and multiplications. The $g + h$ in the first is the addition of the group, the $a + b$ in the second is the addition of the ring, $ag$ is the module multiplication, and the $ab$ in the third is the multiplication of the ring. Just as a computer language can overload an operator and still tell which one is meant from the kinds of its arguments, we can tell by what stands on each side.

Finally, the multiplicative identity of the ring must also act as the identity on the group.

That 1 is the identity of the ring does not make this map respect it automatically, so it is demanded separately.

Q. Doesn’t $1 \cdot g = g$ follow from the three equations above?
A. It does not. If we define every $ag$ to be 0, all three equations hold, yet this map fails $1 \cdot g = g$. (This counterexample is added to the explanation above and was not covered in the lecture.)

A vector space is a special module

In a module, if the ring $R$ is a field $F$ it is a vector space. All that changes is whether the ring has inverses and whether it is commutative. So every vector space is a module, but not every module is a vector space. There is an intermediate stage where the ring is a division ring: with quaternion scalars the old definition counts it as a vector space, but the modern definition requires a field, so it is treated as a module.

The difference is not only in words. A vector space, finite or infinite dimensional, always has a basis, and the proof uses the axiom of choice. A module sometimes has a basis and sometimes does not. (As an example, in $\mathbb{Z}_6$ some elements become 0 when multiplied by an integer, so no basis can be chosen. This example was not covered in the lecture.) The fact that over a field we can divide by a nonzero scalar is what makes all these proofs work.

Vector fields over a ring of functions

An example of a module in general relativity is tensor fields. Each point $p$ of spacetime $M$ has a vector space, the tangent space $T_p$, and a vector field picks one vector at each point. In the coordinate basis $\partial_\mu$ it is $X = X^\mu(x)\, \partial_\mu$, and a vector field is a map sending functions to functions. Vector fields can be added by adding components, so the set of all vector fields is an abelian group.

Now what should we take as scalars? On the set of all infinitely differentiable functions $C^\infty(M)$ we can define addition and multiplication pointwise.

There is also an identity: the function that is 1 at every point. But there are no inverses. A function that takes the value 0 at some point cannot be multiplied pointwise to give 1. So $C^\infty(M)$ is only a ring, neither a division ring nor a field. A vector field times a function is again a vector field, so the set of all vector fields is a module over $C^\infty(M)$. If we restrict the scalars to constant real numbers it is a vector space, but allowing functions makes it only a module.

Q. Why is the set of all functions not a field?
A. A function that passes through a point where it is 0 cannot get a multiplicative inverse. The function $x$ is 0 at $x = 0$, so $1/x$ is undefined at that point. (This specific function is for illustration. The lecture only says a function that takes the value 0 has no inverse.)

Algebra

The most complicated structure is the algebra. Starting from a vector space $V$, we add one more operation that multiplies vectors by vectors.

This product must distribute over addition on both sides and be compatible with the scalars. This is called bilinear.

Here the product $\cdot$ is the product between vectors, and it must be told apart from the product $ab$ of scalars. An algebra is a vector space in which vectors can also be multiplied.

Note that this product need not be associative. Every structure so far demanded associativity first, and in an algebra it becomes optional for the first time. Among the algebras that are interesting in physics, many are non-associative. (The bracket of a Lie algebra and the cross product of 3-vectors are such examples. These examples are for illustration and were not covered in the lecture.) The real and complex numbers are algebras too, and there are many commutative and non-commutative ones.

Building the tensor algebra

The tensor algebra is an important example among these, because it is the foundation for understanding differential forms later. First we review two things. The $n$-fold tensor product of a vector space $V$ is written $V^{\otimes n}$ for short. $V^{\otimes 0}$ is the scalar space itself, $V^{\otimes 1}$ is $V$, and $V^{\otimes 2}$ is the space of rank 2 tensors. And the direct sum of several vector spaces is the set of all ordered tuples $(a, b, c, \dots)$, where the first slot comes from the first space and the second slot from the second space. Addition and scalar multiplication are done slot by slot.

The set of the tensor algebra is the direct sum of the tensor powers of all degrees.

One element is a list holding in turn a real number $a$, a vector $b^\mu e_\mu$, a rank 2 tensor $c^{\mu\nu} e_\mu \otimes e_\nu$, a rank 3 tensor, and so on in the slots. Each slot is a different vector space, and addition and scalar multiplication happen inside the slot. (The standard definition collects only the lists with finitely many nonzero slots. This remark was not covered in the lecture.)

Injection maps and the product

Tensors of different ranks are elements of different vector spaces, so they cannot be added. Yet books sometimes write expressions like $B^{\mu\nu} e_\mu \otimes e_\nu + A^\mu e_\mu$. The device that makes this meaningful is the injection map $\iota_n$. $\iota_n$ puts an element of $V^{\otimes n}$ into the $n$-th slot of $TV$ and fills the rest with 0.

Then $\iota_2(B) + \iota_4(C)$ is an addition inside $TV$, and adding slot by slot gives $(0, 0, B, 0, C, 0, \dots)$. The sum written in books is really a shorthand for the sum taken after this injection map.

Q. We said adding tensors of different ranks is impossible, so how did it become possible?
A. Inside the original spaces it is still impossible. Instead, we move the two into different slots of the larger space $TV$ and use the addition of that larger space. The injection map only places the element as it is, so it creates nothing new.

Multiplication is defined by the tensor product. Taking the tensor product of $B \in V^{\otimes i}$ and $C \in V^{\otimes j}$ gives an element of $V^{\otimes (i+j)}$. With the injection maps this product becomes an operation inside $TV$.

The left side multiplies two elements of $TV$, and the right side is also an element of $TV$. The identity of the product is the list $(1, 0, 0, \dots)$ with 1 in the first slot and 0 elsewhere. For general elements the product is bilinear, so we apply the formula above slot by slot and expand. This is like multiplying polynomials, the difference being that the degrees live in different vector spaces.

Playing with it

The toy below shows this ladder in one place. Pick the scalar ring and see whether you reach a module, a vector space, or an algebra, and in the lower part choose the degrees $i$ and $j$ and watch two tensors be put into slots, added, and multiplied. With quaternion scalars it stops at the old-definition vector space, and with a function ring it stops at a module. Turn off $1 \cdot g = g$ and no ring gives a module. (Examples in this toy such as $\mathbb{Z}_6$ and $\mathbb{Z}_5$ were chosen for illustration and were not covered in the lecture.)

Among the algebras made by taking only some pieces of the tensor algebra, the most important is the exterior algebra, which we will cover soon.