# The “wrong” definitions of “Algebra” and “Lie Algebra”

I am reading Robert Mann's "An Introduction to Particle Physics and the Standard Model" and a review from Amazon (with 12 upvotes) concerns me:

...The definition of algebra on page 54 is wrong. The definition of Lie algebra also on page 54 is wrong. I don't mean typo, I mean wrong...

The definition of Algebra in the book is:

An algebra is a vector space $$V = \{v_I\}$$ with a binary (combining) operation $$\circ$$ that obeys this rule: $$v^I \circ v^J = \sum_{k} C^{IJ}_kv^k = C^{IJ}_kv^k$$, where $$C^{IJ}_k$$ is known as structure constant.

Whilst the definition of Lie Algebra is:

A Lie algebra is an algebra whose combining operation is the commutator: $$\circ = [,]$$ where $$[T^a,T^b] = if^{ab}_cT^c$$ and it is the Taylor-series expansion of a Lie Group. The group associativity implies the Jacobi identity: $$[[T^a,T^b],T^c] + [[T^b,T^c],T^a] + [[T^c,T^a],T^b] = 0$$.

As a novice in the subject, I can't tell what is wrong with these definitions. Can anyone clarify what's wrong with them and give the correct definitions?

• They look about right to me (an algebra can be abstracted further to modules, I believe). Not everyone on the internet is correct. Maybe that includes me. In any case, Lie algebras at a minimum are common enough there's a torrent of sources stating the definition from various points of view (equivalent, but the words aren't always the same) you can compare against. – Richard Myers Mar 16 at 4:24
• I think the definition of a Lie algebra needs some additional constraints on the structure constants, but I'm not sure without checking the details. Other than that I think these definitions are fine, although I can see how a mathematically inclined reader might find them imprecise. – d_b Mar 16 at 4:30
• You can have mathematical fun with the differences between associative and non-associative algebras, algebras over rings and algebras over fields, unitary and non-unitary algebras (or should that be "unital" not "unitary?), etc. Or you can just get on and do some physics. From the title, I guess this book isn't meant for mathematicians to want to quibble over details. – alephzero Mar 16 at 5:17
• You could compare these definitions with those of Wikipedia for algebra and Lie algebra. Mathematicians don’t like thinking of Lie algebras in terms of commutators. $[a,b]$ in a Lie algebra is supposed to be just a “Lie bracket”, a bilinear product producing another element of the algebra; there is no assumption that $ab$ or $ba$ or $ab-ba$ is defined. – G. Smith Mar 16 at 5:23
• I wasn't sure which definition of "algebra" should I look for until @G.Smith gave the relevant Wikipedia page. It's good to know that the definitions above are acceptable from the physics perspective. Appreciate the answers. – cZe99 Mar 16 at 9:16

For a physicist these definitions suffice (unless you get into aspects of theoretical physics where you need to be very precise with your definitions). The complaint is probably from someone with a very mathematical inclination (along with the up votes) but the issues tend not to cause problems in everyday use.

There are two places where I can see this as a legitimate argument coming from:

1. The book never states that the binary operator for the algebra is bilinear, since we are probably going to think of these vector elements as matrices and the binary operator as matrix multiplication this doesn't matter.
2. The book never states that the binary (multiplication/bracket) operator for the Lie algebra will always be anticommutative, $$[x,y]=-[y,x]$$ (note this is anticommutative where $$A$$ times $$B$$ is $$[A,B]$$ rather than $$AB$$). Since we are going to use the commutator $$[x,y]=-[y,x]$$ is obvious.

There is a technical third point that the lie algebra binary operator isn't stated to satisfy $$[x,x] = 0$$. Even if we correct point 2, if the field that the algebra is defined over has field characteristic 2 then we find that $$2a = 0 \quad \forall a$$. This means that while normally $$[x,y] = -[y,x] \Rightarrow [x,x] = 0$$, we instead find that $$[x,x] = -[x,x]$$ but that is true for any element of your lie algebra, not just for zero and so we could have $$[x,x] \neq 0$$. We tend to work over the real or complex numbers, so the field is of characteristic 0 and this point again isn't an issue.

As you can see you probably wouldn't ever run into these because we would only use a small class of algebras in this book, but if you want to try to use more general mathematics for your physics you need to be careful with your definitions. One example is p-adic quantum mechanics or p-adic QFT where instead of your field being the real numbers/complex numbers, your field is the p-adic numbers, i.e. integers mod p. If you want to consider particle physics and the standard model via this method then you start seeing why you need to be more careful. Similar if you want to use algebras where you can't use/don't want to use a matrix representation (I think von Neumann algebras might be a relevant example of this case).