1

Let $f:{\Bbb R}^n\to{\Bbb R}$ be a smooth function. Its differential of $k$-th degree in a point $a\in{\Bbb R}^n$ can be defined as a map $$ d^k f(a):{\Bbb R}^n\to{\Bbb R},\qquad d^kf(a)(p)=\frac{d^k}{d t^k}f(a+tp)\Big|_{t=0},\qquad p\in{\Bbb R}^n. $$ For $k\le 2$ differentials are well-known objects:

  1. For $k=1$ the differential $df(a):{\Bbb R}^n\to{\Bbb R}$ is a linear form (or linear functional) on ${\Bbb R}^n$.

  2. For $k=2$ the differential $d^2f(a):{\Bbb R}^n\to{\Bbb R}$ is a quadratic form on ${\Bbb R}^n$.

Question:

What kind of objects are differentials of degree $k>2$?

There must be analogs of quadratic forms of degree $k>2$, where are they described?

Sergei Akbarov
  • 2,445
  • @Shalop, for $k=2$ there is a theory that establishes a bijection between quadratic forms and symmetric bilinear forms. What is this theory for $k>2$? I even can't find any mentionings of these analogs of quadratic forms. – Sergei Akbarov Apr 03 '15 at 06:14

1 Answers1

3

I guess the classical name is "algebraic form of degree $k$." The next step after quadratic forms is $k = 3$ when you get cubic forms, then quartic forms, and so forth. Algebraic forms of degree $k$ on a finite-dimensional vector space $V$ can be identified with homogeneous polynomials of degree $k$ on $V$, or equivalently with elements of the symmetric power $S^k(V^{\ast})$.

There is a bijection between algebraic forms in this sense and symmetric multilinear forms which concretely is given by polarization and which abstractly reflects a natural isomorphism

$$S^k(V^{\ast}) \cong S^k(V)^{\ast}$$

(say, in characteristic zero).

The higher derivatives of a smooth function $f : V \to \mathbb{R}$ are just the coefficients of its Taylor polynomials. Taylor polynomials, like all polynomials, break up into a sum of homogeneous polynomials, one in each degree. The degree zero part is $f(0)$, the degree one part encodes the first derivatives, etc. (Here I am always taking derivatives at $0$ WLOG.)

For example, in two variables the Taylor expansion takes the form

$$f(x, y) = \sum_n^{\infty} \sum_{i + j = n} f_{ij} x^i y^j$$

where

$$f_{ij} = \frac{1}{i! j!} \frac{\partial^i f}{\partial x^i \partial y^j} \Big|_{x=y=0}.$$

So e.g. the constant part is $f_{00}$, the linear part is $f_{10} x + f_{01} y$, the quadratic part is $f_{20} x^2 + f_{11} xy + f_{02} y^2$, the cubic part is $f_{30} x^3 + \dots$ and hopefully you get the idea. And of course you can isolate the degree $k$ part by looking at the coefficient of $t^k$ in $f(tx, ty)$.

Qiaochu Yuan
  • 419,620
  • And for the global story see http://en.wikipedia.org/wiki/Jet_bundle. – Qiaochu Yuan Apr 03 '15 at 08:56
  • "Algebraic form of degree $k$"? – Sergei Akbarov Apr 03 '15 at 08:59
  • @Sergei: well, you could also just say "homogeneous polynomial of degree $k$." It depends on how classical you want to be, I guess. – Qiaochu Yuan Apr 03 '15 at 09:01
  • Qiaochu, I think it's not good to refer to books on jet bundles when mentioning these elementary facts. Do you know a textbook on algebra where this bijection between homogenious polynomials and symmetric multilinear forms is described? – Sergei Akbarov Apr 03 '15 at 09:07
  • @Sergei: possibly the reference given in the Wikipedia article. I haven't looked at it myself. In any case, you were talking about higher derivatives of smooth functions and it seems like a natural follow-up question. – Qiaochu Yuan Apr 03 '15 at 09:11
  • I need a reference... It's not good to invent a bicycle... – Sergei Akbarov Apr 03 '15 at 09:13
  • @Sergei: just look at the reference given in the Wikipedia article. It's Procesi, Lie Groups: an approach through invariants and representations. Chapter 3 is devoted to algebraic forms, and polarization is section 2 in that chapter. – Qiaochu Yuan Apr 03 '15 at 09:17
  • It will take me some time to find this book... – Sergei Akbarov Apr 03 '15 at 10:03
  • What I don't understand in this science: if quadratic forms have analogs in higher degrees, then there must be analogs of parallelogram identity for cubic forms, quartic froms, etc. What are they? – Sergei Akbarov Apr 04 '15 at 12:52
  • @Sergei: they are given by polarization. – Qiaochu Yuan Apr 04 '15 at 19:12
  • Qiaochu, I feel like a student with you. How does this polarization define "parallelogram identity", say, for cubic forms? – Sergei Akbarov Apr 04 '15 at 21:35
  • @Sergei: sorry, I guess it's a little more complicated than just polarization. You want to isolate the coefficient of $x_1 x_2 \dots x_k$ in $f(x_1 v_1 + x_2 v_2 + \dots x_k v_k)$. In general you should be able to do it by substituting suitable roots of unity for the $x_i$ and taking a suitable linear combination. I don't have details off the top of my head though. – Qiaochu Yuan Apr 05 '15 at 04:33
  • OK, I posted this as another question: http://math.stackexchange.com/questions/1226024/analogs-of-the-paralleloram-identity-in-higher-degrees – Sergei Akbarov Apr 09 '15 at 06:11