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From Humphreys' Introduction to Lie Algebras and Representation Theory:

By an $F$-algebra (not necessarily associative) we simply mean a vector space $U$ over $F$ endowed with a bilinear operation $U\times U\rightarrow U$, usually denoted by juxtaposition (unless $U$ is a Lie algebra, in which case we always use the bracket). By a derivation of $U$ we mean a linear map $\delta:U\rightarrow U$ satisfying the familiar product rule $\delta(ab)=a\delta(b)+\delta(a)b$.

I'm wondering what is an example of a derivation. Suppose I take $U=\mathbb{R}^n$. Clearly the map that takes everything to $0$ is a derivation. The map $\delta(x)=kx$ is not a derivation for $k\neq 0$, because then $kab\neq kab+kab$. What are some other linear maps satisfying that product rule?

PJ Miller
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  • What $\mathbb{R}$-algebra structure do you impose on $\mathbb{R}^n$? By the way, this is a great question; it's always important to question things in mathematics and find examples of your own! – Amitesh Datta Aug 10 '13 at 03:21

1 Answers1

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As I commented above, I'm not sure what $\mathbb{R}$-algebra structure you're referring to on $\mathbb{R}^n$. The cross product endows $\mathbb{R}^{3}$ with the structure of an $\mathbb{R}$-algebra but this isn't applicable in higher dimensions. You can, however, consider the quaternions, the octonions etc. as these furnish $\mathbb{R}$-algebras. The following answer provides some examples of derivations and furnishes some practice with the notion.

The polynomial ring $\mathbb{R}[x]$ is an $\mathbb{R}$-algebra. A derivation $\delta:\mathbb{R}[x]\to \mathbb{R}[x]$ is given by the rule $\delta(f)=\frac{df}{dx}$. (In fact, this example is the initial motivation for the definition of "derivation".) The set of derivations of an $\mathbb{F}$-algebra is an $F$-vector space. In particular, any scalar multiple of $\delta$ is again a derivation in the example of the first two sentences of this paragraph.

In general, a derivation $\delta$ of an $F$-algebra $U$ has the following properties:

(1) $\delta(1)=0$ if $1$ is the unity of $U$.

(2) $\delta(u^n)=n\delta(u^{n-1})$ for all $u\in U$.

Exercise 1: Prove (1) and (2).

Exercise 2: Classify all derivations of the $\mathbb{R}$-algebra $\mathbb{R}[x]$. (Hint: A derivation $\delta$ of $\mathbb{R}[x]$ is determined by $\delta(x)$. Why? Also, use Exercise 1.)

I hope this helps and provides some practice with the concept of a derivation!

Amitesh Datta
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  • Thanks, Amitesh! I've read up to your third paragraph, and have a couple of questions:

    (1) $U$ is a vector space. How do we define its unity?

    (2) Why can we write $u^n$? Do we know that multiplication in $U$ is associative?

     – PJ Miller Aug 10 '13 at 04:19
  • Hi @PJMiller, an $F$-algebra $U$ is an $F$-vector space with the structure of a ring. The two structures are required to be compatible so that if $u,v\in U$ and $f\in F$, then $f(uv)=(fu)v=u(fv)$ (notice that I've abused notation and denoted multiplication by the scalar ($f$) as well as multiplication in $U$ by juxtaposition). In an $F$-algebra, we can always speak of a unity because an $F$-algebra is a ring and a vector space. – Amitesh Datta Aug 10 '13 at 05:03
  • I'd just like to add that the structure of an $F$-vector space on a ring (i.e., an $F$-algebra) is quite rich. Not every integral domain is a field. (E.g., the ring of integers, $\mathbb{Z}$!) However, every $F$-algebra which is also an integral domain must be a field. Why? Well, if $x\in U$ where $U$ is an $F$-algebra, then the map $\phi_x:U\to U$ defined by the rule $\phi_x(u)=xu$ ($xu$ denotes the product of $x$ and $u$ in $U$) is an $F$-linear map. It's injective because $U$ is an integral domain and so must be surjective by rank-nullity. Therefore, there exists $y\in U$ such that $xy=1$! – Amitesh Datta Aug 10 '13 at 05:07
  • Yes, @PJMiller, we do know that multiplication in $U$ is associative just like in any ring. (You should think of an $F$-algebra as a ring which is also a vector space: freely multiply and don't worry about simple things like associativity and existence of unity. You can talk about ideals, subalgebras, quotient algebras etc. in an $F$-algebra just like you can do so in a ring!) – Amitesh Datta Aug 10 '13 at 05:09
  • Thanks, Amitesh! One more question: from the first sentence of the text I quoted "By an $F$-algebra (not necessarily associative) we simply mean a vector space $U$ over $F$ endowed with a bilinear operation $U\times U\rightarrow U$". What does "not necessarily associative" mean here? Since you said multiplication in $U$ is associative. – PJ Miller Aug 10 '13 at 06:05
  • Hi @PJMiller, I didn't notice that and you're right! I was considering associative algebras in my answer but nonassociative algebras are also fundamental. E.g., see http://en.wikipedia.org/wiki/Nonassociative_algebra. In order to define $u^n$ unambiguously for $u\in U$, one needs only the weaker condition of power associativity. I can't think of any useful instances of considering non power associative algebras in mathematics at the moment though that's more a reflection of my ignorance than of mathematics. – Amitesh Datta Aug 10 '13 at 07:19