Are there are different types of algebra?

Question

I've begun studying an introductory linear algebra textbook. In the first sections, all of the content focuses on defining a vector space over a field. In doing so, the author emphasises defining each valid operation for a given vector space, as well as the valid scalars and operations with scalars as defined by the field.

So far, it seems that all of the operations with vector spaces and scalars are identical to what has been taught during more basic mathematics courses (high school). However, the author seems to imply that you can define the algebraic operations that most people find "conventional" in completely arbitrary ways.

Despite learning mathematics using one type of algebra throughout earlier courses, is it actually normal for mathematicians to create arbitrary types of algebra, with operations such as addition and multiplication defined in "unconventional" ways? If so, why?

Is the "conventional" algebra taught in earlier courses more widespread because it models the real world, unlike "unconventional" algebras?

Answer 1

Yes, completely.

First of all, there's a theorem that says that any finite dimensional vector space over a field $\mathbb F$ is isomorphic to $\mathbb F^n$ for some power of $n$. So why study generic finite dimensional vector spaces anyway? Why not focus on $\mathbb F^n$? This answer goes into that.

Now, a natural question in dealing with vector spaces is why must we use a field? You might recall that $\mathbb R$ is a field, $\mathbb Q$ is a field, but $\mathbb Z$ isn't. Can we define "Vector spaces" over $\mathbb Z$? We can multiply by things in $\mathbb Z$, so it seems like there should be some way of doing it. The answer is that we can, and they're called modules, or specifically a module over a ring. A ring is like a field, except:

1. It doesn't have to be commutative (sets of $n\times n$ matrices are rings, and $AB\neq BA$ for matrices $A,B$).

2. It doesn't have to contain an identity element ($2\mathbb Z$ is the set of all even numbers. You can add them and you get another even number, you can multiply them together and you get another even number. But there isn't any element that's the multiplicative identity).

3. You can have zero divisors, or elements $A\neq 0, B\neq 0$ such that $AB = 0$. Again, matrices are a good example here, specifically consider: $$A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ Then, $A^2 = 0$, but $A\neq 0$.

4. You don't have to have (multiplicative) inverses. Again, think of $2\mathbb Z$ - none of it's elements have multiplicative inverses (there isn't even an identity element).

A module is just one of many different algebraic structures we can study. You might have noticed I mentioned the word "ring" above, and compared it to a field. A ring is another algebraic structure we can study, and a field is just a very special kind of ring.

The other "fundamental" (meaning typically introduced at an undergraduate level) object of study is a group. For rings, for something like $2\mathbb Z$, to be overly pedantic I should really write $(2\mathbb Z,+,\times)$, as a ring is a set of elements, along with two binary operations that satisfy some rules (associativity, distributivity, etc.). A group is a set along with a single binary operation that's associative, has an identity, and all elements are invertible. So, we could look at $(\mathbb Z,+)$ as a group, or even something like $(C([0,1]),+)$ (continuous functions on the interval $[0,1]$).

All of these have a ton of interplay. There's a result that says a $\mathbb Z$-module (so, a "vector space" over $\mathbb Z$) is really just the same thing as a group, and vice versa. Additionally, given some ring (such as $n\times n$ matrices), we can look at some "fundamental" groups associated with it, like the multiplicatively invertible elements.

For a field, everything nonzero has an inverse, so $\mathbb Q^\times$ (the $\times$ means the invertible elements) is the same as $\mathbb Q\setminus\{0\}$, where these are both groups under multiplication. More interestingly, we can start with $n\times n$ matrices, and look at the multiplicative group of these, which is called $GL_n(\mathbb F)$, or the general linear group. As you may know now, or learn in your linear algebra course, a matrix $A$ is invertible if and only if $\det A \neq 0$. This is also the "defining condition" for a matrix to be in $GL_n(\mathbb F)$!

Answer 2

There are a dizzying array of different types of algebraic systems (groups, rings, modules, vector spaces, algebras, fields, ...).

Yes, mathematicians sometimes create "unconventional" things out of thin air. Why? Often, it's to answer a question such as "Does there exist a ... with properties ... and ..., but not ...?".

So, why should any sane person care about that? Well, for one thing, it turns out that lots of these mathematical structures keep turning up in applications, and knowing about the properties of those structures gives you real insight into the applications. For example, vector spaces over finite fields, which 100 years ago people would have thought to be hopelessly abstract, turns out to be just what you need to study error-correcting codes.

Answer 3

Consider any vector space $V$, and lets assume, for simplicity, that is has a finite basis $v_1,\ldots,v_n$. To define an algebra structure on $V$, we must assign to each pair of vectors $v,w$ a product which I will denote by $vw$.

Because this product is supposed to be distributive with respect to addition, and because the vectors $v_1,\ldots,v_n$ form a basis for $V$, we see it suffices to define the products $v_iv_j$ for all possible pairs $(i,j)$. We skipped another detail: we also want that the product behaves well with scalar multiplication, that is, if $\lambda$ is a scalar, we want that $\lambda (vw)= (\lambda v)w = v(\lambda w)$. This is what justifies the claim above, along with the distributive property.

Now each product $v_iv_j$ is also a vector, so we can write it as a linear combination $$\sum_{k=1}^n \lambda_{ij}^k v_k$$

for scalars $\lambda_{ij}^k$. We call the $n^3$ scalars the structure constants of this putative multiplication.

Because we want multiplication to be associative, the equalities $v_i(v_jv_k)=(v_iv_j)v_k$ need to hold true. This imposes certain polynomial relations among these scalars (try to write down some for small $n$), and the existence of a unit element $1 = \sum \mu_k v_k$ imposes relations among the scalars $\mu_k$ and the chosen structure constants.

Moreover, any finite dimensional algebra arises in this way: if $A$ is a $k$-algebra and if $v_1,\ldots,v_n$ is a basis of $A$, calculate the products $v_iv_j$ to obtain the structure constants. For example, consider the vector space $M_n(k)$ of matrices of size $n$ over $k$. This has basis the elementary matrices $e^{ij}$ with a $1$ in entry $(i,j)$ and a zero elsewhere, and the structure constants are Kronecker deltas.