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.

From Milnor and Stasheff:

If the $n$-dimensional manifold $M$ can be immersed in $\mathbb{R}^{n+1}$ show that each Stiefel-Whitney class $w_i(M)$ is equal to the $i$-fold cup product $w_1(M)^i$.

From the Whitney duality theorem we immediately have that the dual Stiefel-Whitney classes $\bar{w_i}(M)$ are zero for $i>1$. But I don't see how to use this to show that each class is a cup product of the first Stiefel-Whitney class.

Looking at the second half of the question which is to show that $\mathbb{R}P^n$ can be immersed in $\mathbb{R}^{n+1}$ only if $n=2^r-1$ or $2^r-2$ is not hard - the dual total Stiefel-Whitney classes can only be 1 or $1+a$. In the first case we get that the total Stiefel-Whitney classes must also be 1, which is true when $n=2^r-1$, and in the second we find $w(M)=1+a+a^2+ \cdots + a^n$, which is true if $n=2^r-2$ (I think that all works).

Is there any way to salvage an argument for the first part of the question from this? Indeed for a codimension 1 immersion the dual total Stiefel-Whitney class can only be 1 or $1+\bar{w_1}(M)$. But I can't see how this implies that $w_i(M) = w_1(M)^i$.

Any hints/tips?

Edit I think I actually had it written above! I really only need to worry about the case where $\bar{w_1}(M) \ne 0$, else all the non-zero Stiefel-Whitney classes are zero. So I can construct the formal inverse of $1+\bar{w_1}(M)$ which is given by $$ \begin{eqnarray*} w_1 &=& \bar{w_1} \\ w_2 &=& \bar{w_1}^2+\bar{w_2} = w_1^2 \\ w_3 &=& \bar{w_1}^3+\bar{w_3}=w_1^3 \end{eqnarray*} $$ and in general $w_i(M) = w_1(M)^i$

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

From the immersion $i:M\to \mathbb R^n$ you get the exact sequence of bundles on $M$ (in which $N$ denotes the normal bundle of the immersion)
$$0\to T(M)\to i^\ast(T\mathbb R^n)\to N\to 0 $$ Since the middle bundle is trivial you have (using Whitney multiplicativity)
$$w(T(M)).w(N)=w(i^\ast(T\mathbb R^n)=1 $$ so that [writing as usual $w(M)$ for $w(T(M))$] $$w(M)=w(N)^{-1}=(1+w_1(N))^{-1}=1+w_1(N)+w_1(N)^2+\ldots+w_1(N)^{i}+\ldots (*)$$ and so $w_i(M)=w_1(N)^i (**) \;$ , which is not yet quite the result. However, if you consider the component of degree one in the graded equality $(\ast)$ , you'll get $w_1(M)=w_1(N)$ and so finally putting this into $(**)$ , you obtain $w_i(M)=w_1(M)^i$, the desired equality.

share|improve this answer
    
Dear Georges, thank you very much. I believe this is essentially what I ended up working out above. –  Juan S Jul 13 '11 at 8:55
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.