Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

INTRODUCTION

Let $R$ be any ring, possibly without unity. We call a function $d:R\longrightarrow R$ a derivation on $R$ if it satisfies the following conditions.

$(1)$ It is an endomorphism of the additive abelian group of $R;$

$(2)$ For any $r,s\in R$ we have that $d(rs)=d(r)s+rd(s).$

I believe that when $R$ is an $A$-algebra for some commutative ring $A,$ we should also make the assumption that $d$ respects scalar multiplication, but I haven't seen the definition other than in the case of $R=K[x]$ and $A=K$ for a field $K.$

In this case, more specifically for $K=\mathbb C,$ it is standard knowledge that the standard derivative satisfies these conditions. Actually, we could just as well take the algebra of all holomorphic functions (instead of polynomials only) with the standard derivative.

However, there are more such functions. It was a homework assignment in my field theory class to prove that for any $f\in K[x]$ there is exactly one derivation $d_f:K[x]\longrightarrow K[x]$ such that $d_f(x)=f.$ This is an easy exercise and one easily sees that in particular, for $K=\mathbb C,$ the standard derivative is just $d_1.$

QUESTION

I understand that I will soon be shown how such derivations can be used in algebra, and I can't wait to see that. But since it's a field theory class, I'm quite sure that nothing will be said about their meaning outside algebra. This is why I'm writing this.

I would like to know

$(a)$ whether the derivations $d_f$ on $K[x]$ I wrote about above have any interpretation connected to actually differentiating something (which I understand involves taking limits in some metric space) and what the interpretation is. In particular, can we meaningfully assign a metric space to an arbitrary field?

$(b)$ whether the general definition of a derivation on a ring or algebra has such an an interpretation.

share|improve this question
3  
The derivation $d_f$ on $K[x]$ that sends $x$ to $f$ is not a mystery at all: it's just $f(x)d/dx$. That is, $d_f(g) = fg'$. I think you must have solved that homework problem without even realizing what your answer was, concretely. The point, then, is that if $d \colon R \rightarrow R$ is a derivation and $a \in R$, then $ad$ is also a derivation, where $(ad)(r) = a\cdot d(r)$. It's similar to a scalar multiple of a linear map also being a linear map. –  KCd Mar 24 '12 at 16:21
3  
Find a book on differential geometry that discusses tangent spaces to see a role for derivations in a geometric setting. –  KCd Mar 24 '12 at 16:25

2 Answers 2

up vote 4 down vote accepted

Let $M$ be a differential manifold of class $\mathcal C^\infty$ and $m\in M$ be a point.

1) A very elementary and efficient way to define the tangent space $T_m(M)$ is to say that it is the space of derivations of the algebra of germs of smooth functions defined in a neighbourhood of $m$:
$$T_m(M)=Der_{\mathbb R}(\mathcal C^\infty _m,\mathbb R)$$
2) In a similar vein the space of global vector fields of $M$ can be identified with the set of derivations of the algebra of global differentiable functions into itself: $$\mathcal X(M)=\Gamma(M,TM)=Der_{\mathbb R} (C^\infty (M),C^\infty (M) )$$

3) In differential geometry there are alternative ways of defining the tangent space $T_m(M)$: by equivalence classes of curves through $m$ or by giving a huge collection of vectors, one for each chart around $m$ , related by linear maps associated to each change of coordinates.
This last point of view was popularized by Einstein and other physicists more than a century ago.

4) However in algebraic geometry, especially when studying singular varieties, derivations (or their variants like Zariski tangent space) are essentially the only tool for defining tangent spaces of varieties.
For example, if you consider the cusp $M\subset \mathbb A^2_k$ with equation $y^2-x^3=0$, its tangent space at the origin $m=(0,0)$ is the vector space $T_m(M)=k\cdot \frac {\partial} {\partial x} \oplus k\cdot \frac {\partial} {\partial y}$.
This is weird since it is a $2$- dimensional vector space whereas $M$ is $1$-dimensional.
This phenomenon, impossible for differential manifolds, is due to the singularity of $M$ at $m$.

share|improve this answer

A derivation on an algebra $A$ is an infinitesimal automorphism, and the collection of derivations on $A$ naturally forms a Lie algebra which one should think of as the "Lie algebra of the automorphism group of $A$." When $A = C^{\infty}(\mathbb{R})$, the automorphism group corresponding to ordinary differentiation is translation $\alpha_t : f(x) \mapsto f(x+t)$, and more generally when $A = C^{\infty}(M)$ for $M$ a compact smooth manifold, a derivation on $A$ is the same thing as a vector field on $M$.

Not all derivations actually correspond to automorphisms. Given a derivation $D : A \to A$, formally the corresponding one-parameter family of automorphisms is supposed to be $$e^{Dt} = \sum_{n \ge 0} D^n \frac{t^n}{n!}$$

and formally taking the derivative of this family recovers $D$, but the family usually only exists formally as a map $A \to A[[t]]$. However, if $A$ has extra structure then the above map may actually exist, for example if $A$ is a Banach algebra.

So derivations not only appear in algebra, but also in differential geometry and functional analysis. A longer explanation of what I just said above may be found at this blog post.

share|improve this answer
1  
Dear @Qiaochu, what you write is correct but I'd like to mention that a derivation of $A$ is the same thing as a vector field on $M$ also if $M$ is not compact. –  Georges Elencwajg Mar 24 '12 at 17:34
    
@Georges: interesting! I was under the impression that the space of characters $C^{\infty}(M) \to \mathbb{R}$ failed to be homeomorphic to $M$ if we didn't assume compactness, so I expected a similar thing to be true about derivations. –  Qiaochu Yuan Mar 24 '12 at 17:36
    
Thank you very much. This is beyond my comprehension now, but I will surely come back to this. –  user23211 Mar 25 '12 at 23:39

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.