# The definition of $s_I$ in Milnor and Stasheff, Characteristic classes, page 188

I cannot understand the definition of polynomials $$s_I$$ in the Milnor and Stasheff, Characteristic classes, page $$188.$$

In Milnor and Stasheff, Characteristic classes page $$188$$, the polynomials $$S_I$$ are defined as follows.

Let $$t_1,...,t_n$$ be indeterminates.

Now for any partition $$I=i_1,...,i_r$$ of $$k$$, define a polynomial $$S_I$$ in $$k$$ variables as follows. Choose $$n \geq k$$ so that the elementary symmetric functions $$\sigma_1,...,\sigma_k$$ of $$t_1,...t_n$$ are algebraically independent and let $$s_I(\sigma_1,....,\sigma_k)= \sum t_1^{i_1}...t_r^{i_r}$$.

The summation $$s_I(\sigma_1,....,\sigma_k)= \sum t_1^{i_1}...t_r^{i_r}$$ is taken over all monomials transformed from $$t_1^{i_1}...t_r^{i_r}$$ by the permuation group acting on the set $$\{t_1,....,t_n\}$$.

$$s() = 1$$

$$s_1(\sigma_1) = \sigma_1$$

$$s_{2}(\sigma_1,\sigma_2) = \sigma_1^2-2\sigma_2$$

$$s_{1,1}(\sigma_1,\sigma_2) = \sigma_2$$

$$s_{3}(\sigma_1,\sigma_2,\sigma_3) = \sigma_1^3 - 3 \sigma_1 \sigma_2 + 3\sigma_3$$

$$s_{1,2}(\sigma_1,\sigma_2,\sigma_3) = \sigma_1 \sigma_2 - 3\sigma_3$$

$$s_{1,1,1}(\sigma_1,\sigma_2,\sigma_3) = \sigma_3$$

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Edit after the answer of Eric

I can understand the above equations. For example, consider the following equation.

$$s_{3}(\sigma_1,\sigma_2,\sigma_3) = \sigma_1^3 - 3 \sigma_1 \sigma_2 + 3\sigma_3$$

This equation means the following.

$$s_3(t_1,t_2,t_3) := (t_1)^3+(t_2)^3+(t_3)^3 \\ = (t_1+t_2+t_3)^3 -3(t_1+t_2+t_3)(t_1t_2+t_2t_3+t_3t_1) -3t_1t_2t_3 \\ = \sigma_1^3 - 3 \sigma_1 \sigma_2 - 3\sigma_3$$

The last sign is minus instead of plus ... I guess the book misprinted?

• Can you elaborate on what exactly you don't understand? Commented Jan 1, 2020 at 22:21
• In particular there are quite a few typos in your examples at the end which might be related to why you are confused. Commented Jan 1, 2020 at 22:38
• In page 188 of Milnor and Stasheff, it says that "Choose $n \geq k$ so that the elementary symmetric functions $\sigma_1,...,\sigma_k$ of $t_1,...t_n$ are algebraically independent ". But I cannnot understand what this means? Commented Jan 2, 2020 at 8:30

Let $$S$$ be the subring of $$\mathbb{Z}[t_1,\dots,t_n]$$ consisting of polynomials that are symmetric in the $$n$$ variables (i.e., invariant under any permutation of the variables). It is a theorem that $$S$$ itself is a polynomial ring in the elementary symmetric polynomials $$\sigma_1,\dots,\sigma_n$$, where $$\sigma_k$$ is the sum of all the distinct monomials that can be obtained from $$t_1\dots t_k$$ by permuting the variables. In other words, every element of $$S$$ can be uniquely written as a polynomial in $$\sigma_1,\dots,\sigma_n$$.

So, here is what the definition of $$s_I$$ means. Take the polynomial $$f(t_1,\dots,t_n)$$ in $$n$$ variables given by summing up all distinct monomials that can be obtained from $$t_1^{i_1}\dots t_r^{i_r}$$ by permuting the variables. Then $$f\in S$$, so it can be written as a polynomial in $$\sigma_1,\dots,\sigma_n$$. In fact, it turns out that $$f$$ is a polynomial in just $$\sigma_1,\dots,\sigma_k$$, and that moreover the coefficients of this polynomial in $$\sigma_1,\dots,\sigma_k$$ do not depend on the choice of $$n\geq k$$. We write $$s_I$$ for the unique polynomial in $$k$$ variables such that $$s_I(\sigma_1,\dots,\sigma_k)=f(t_1,\dots,t_n)$$.

Let's look at an example. For instance, take $$I$$ to be the partition $$(2)$$ and $$n=2$$, so $$f(t_1,t_2)=t_1^2+t_2^2$$. To find $$s_I$$, we want to write $$t_1^2+t_2^2$$ in terms of $$\sigma_1=t_1+t_2$$ and $$\sigma_2=t_1t_2$$. To do that, we can simply observe that $$(t_1+t_2)^2-2t_1t_2=t_1^2+t_2^2$$, so $$f=\sigma_1^2-2\sigma_2$$. Thus the polynomial $$s_I$$ is $$s_I(x_1,x_2)=x_1^2-2x_2$$, since if we evaluate this polynomial at $$(x_1,x_2)=(\sigma_1,\sigma_2)$$ we get $$f$$.

• Thank you. I could not understand with the book, but in your explanation, I can understand it. Thank you. Commented Jan 2, 2020 at 23:17
• How do we know that the coefficients of the polynomial does not depend on the choice of $n$? Commented Aug 20, 2020 at 11:39