4 Another typo. The last one, I hope.

This is not an answer to your question. Rather, it is a "less mysterious" version of the Milnor-Stasheff construction of the Stiefel-Whitney classes, which doesn't refer explicitly to Steenrod operations. (I think I learned about this from my thesis advisor when I was a lad ... it was so long ago ...)

Let $V\to X$ be a real vector bundle. Let $S^\infty=\bigcup S^n$, the infinite dimension sphere. Taking product with $S^\infty$ gives a vector bundle $V\times S^\infty\to X\times S^\infty$. I produce a vector bundle $V'\to X\times RP^\infty$ by dividing out by an action of the cyclic group of order $2$ on both base and total space:

• on the base $X\times S^\infty$, the involution is $(x,y)\mapsto (x,-y)$;
• on the total space $V\times S^\infty$, the involution is $(v,y)\mapsto (-v,-y)$.

The Euler class $e(V')$ of $V'$ is an element of degree $n$ in $H^*(X\times RP^\infty; Z/2) = H^*(X;Z/2)[t]$. The following formula holds: $$e(V') = t^n + w_1(V)t^{n-1}+\cdots + w_n(V).$$ So if you have an Euler class, then you can use this as the definition of the Stiefel-Whitney classes. The mod-2 Euler class is fairly easy to define from Milnor-Stasheff's point of view: $e(V')$ is the pullback along the $0$ section of the orientation class in the cohomology of the Thom space of $V'$.

It's easy to check the axioms for this guy. It's certainly natural, since $V\mapsto V'$ and $e$ are functorial. Whitney sum follows from $(V\oplus W)'\approx V'\oplus W'$ and the Whitney sum formula for the Euler class. If $R\to *$ is the trivial bundle, then $R'\to RP^\infty$ is the canonical line, so $e(R')=t$ and so $w_0(R)=1$ and $w_1(R)=0$. You can use this to show that $w_0(V)\in H^0X$ is equal to $1$ for any bundle over $X$, by pulling back $V$ over any point of $X$ (where it becomes trivial). If $L\to RP^\infty$ is the canonical line, then $L'\to RP^\infty\times RP^\infty$ is $L_1\otimes L_2$, the tensor product of the canonical line bundles over each factor. So $e(L')=s+t= 1\cdot t^1 + s\cdot t^0$, giving $w_0(L)=1$ and $w_1(L)=s$ (where $s\in H^1RP^\infty$ is the generator).

Added later. I wrote the above while I was a bit feverish :). It didn't occur to me when describing it that it's a pretty standard way to construct characteristic classes; the variant which gives chern classes is probably more familiar.

I also said that it's a "version" of the Steenrod operation construction of SW classes, so let me try to explain that. I'll sketch a "direct" proof that the Steenrod operation definition of SW classes is equivalent to the one I gave above (i.e., without refering the axioms that M-S give for SW classes).

Steenrod operations come from an "extended square" construction on cohomology classes (see my answer in http://mathoverflow.net/questions/6377/why-does-one-think-to-steenrod-squares-and-powers/6384#6384). If $X$ is a space, let $DX=(X\times X \times S^\infty)/(Z/2)$, where I divide by the involution $(x_1,x_2,y)\to (x_2,x_1,-y)$. The "extended square" is a function $$P: H^n(X) \to H^{2n}(DX).$$ Cohomology is with mod-2 coefficients. If you restict along the "diagonal" embedding $d: X\times RP^\infty \to DX$, you get Steenrod squares: $$d^*(P(a)) = t^{n}Sq^0(a) + t^{n-1}Sq^1(a) + \cdots + Sq^n(a).$$ There's a relative version of this: if $V\to X$ is a vector bundle, so is $DV\to DX$; write $T(V)$ for the Thom space of $V$, and write $f: T(V) \to T(DV)$ for the map induced by diagonal inclusion. If $u\in H^nT(V)$ is the orientation class, then $$f^*(P(u))= t^{n}Sq^0(u)+t^{n-1}Sq^1(u)+\cdots +Sq^n(u).$$ According to Milnor-Stasheff, $Sq^i(u)=w_i(V)$.Sq^i(u)=u\,w_i(V)$. The neat fact is that$P(u)\in H^{2n}T(DV)$has to be the orientation class$u'$of$DV\to DX$! So as long as I can describe the orientation class, I don't need to know about Steenrod opeartions! Thus,$f^*(u')\in H^*TV[t]$is the polynomial whose coefficients are the SW classes. To get the formula I gave originally, observe that$d^*(u')=u\, f^*(u')=u\, e(V')$; this is because the pullback of the bundle$TV\to TX$along$d: X\to DX$is the same as the bundle$V+V' \to X$. Why is$P(u)$the orientation class of$DV$? The orientation class of a bundle in ordinary cohomology mod-2 is the unique element which restricts to the fundamental class of the sphere when you restrict to each fiber, so you just have to check that$P(u)$has this property. And this is pretty easy (the operation$P$is natural, and it's easy to understand how$P$works when you have a discrete space, or a bundle over a discrete space.) 3 Added a new section, proving that M-S SW classes are the same as these Added later. I wrote the above while I was a bit feverish :). It didn't occur to me when describing it that it's a pretty standard way to construct characteristic classes; the variant which gives chern classes is probably more familiar. I also said that it's a "version" of the Steenrod operation construction of SW classes, so let me try to explain that. I'll sketch a "direct" proof that the Steenrod operation definition of SW classes is equivalent to the one I gave above (i.e., without refering the axioms that M-S give for SW classes). Steenrod operations come from an "extended square" construction on cohomology classes (see my answer in http://mathoverflow.net/questions/6377/why-does-one-think-to-steenrod-squares-and-powers/6384#6384). If$X$is a space, let$DX=(X\times X \times S^\infty)/(Z/2)$, where I divide by the involution$(x_1,x_2,y)\to (x_2,x_1,-y)$. The "extended square" is a function$$P: H^n(X) \to H^{2n}(DX).$$Cohomology is with mod-2 coefficients. If you restict along the "diagonal" embedding$d: X\times RP^\infty \to DX$, you get Steenrod squares:$$d^*(P(a)) = t^{n}Sq^0(a) + t^{n-1}Sq^1(a) + \cdots + Sq^n(a).$$There's a relative version of this: if$V\to X$is a vector bundle, so is$DV\to DX$; write$T(V)$for the Thom space of$V$, and write$f: T(V) \to T(DV)$for the map induced by diagonal inclusion. If$u\in H^nT(V)$is the orientation class, then$$f^*(P(u))= t^{n}Sq^0(u)+t^{n-1}Sq^1(u)+\cdots +Sq^n(u).$$According to Milnor-Stasheff,$Sq^i(u)=w_i(V)$. The neat fact is that$P(u)\in H^{2n}T(DV)$has to be the orientation class$u'$of$DV\to DX$! So as long as I can describe the orientation class, I don't need to know about Steenrod opeartions! Thus,$f^*(u')\in H^*TV[t]$is the polynomial whose coefficients are the SW classes. To get the formula I gave originally, observe that$d^*(u')=u\, e(V')$; this is because the pullback of the bundle$TV\to TX$along$d: X\to DX$is the same as the bundle$V+V' \to X$. Why is$P(u)$the orientation class of$DV$? The orientation class of a bundle in ordinary cohomology mod-2 is the unique element which restricts to the fundamental class of the sphere when you restrict to each fiber, so you just have to check that$P(u)$has this property. And this is pretty easy (the operation$P$is natural, and it's easy to understand how$P$works when you have a discrete space, or a bundle over a discrete space.) 2 fixed typos This is not an answer to your question. Rather, it is a "less mysterious" version of the Milnor-Stasheff construction of the Stiefel-Whitney classes, which doesn't refer explicitly to Steenrod operations. (I think I learned about this from my thesis advisor when I was a lad ... it was so long ago ...) Let$V\to X$be a real vector bundle. Let$S^\infty=\bigcup S^n$, the infinite dimension sphere. Taking product with$S^\infty$gives a vector bundle$V\times S^\infty\to X\times S^\infty$. I produce a vector bundle$V'\to X\times RP^\infty$by dividing out by an action of the cyclic group of order$2$on both base and total space: • on the base$X\times S^\infty$, the involution is$(x,y)\mapsto (x,-y)$; • on the total space$V\times S^\infty$, the involution is$(v,y)\mapsto (-v,-y)$. The Euler class$e(V')$of$V'$is an element of degree$n$in$H^*(X\times RP^\infty; Z/2) = H^*(X;Z/2)[t]$. The following formula holds: $$e(V') = t^n + w_1(V)t^{n-1}+\cdots + w_n(V).$$ So if you have an Euler class, then you can use this as the definition of the Stiefel-Whitney classes. The mod-2 Euler class is fairly easy to define from Milnor-Stasheff's point of view:$e(V')$is the pullback along the$0$section of the orientation class in the cohomology of the Thom space of$V'$. It's easy to check the axioms for this guy. It's certainly natural, since$V\mapsto V'$and$e$are functorial. Whitney sum follows from$(V\oplus W)'\approx V'\oplus W'$and the Whitney sum formula for the Euler class. If$R\to *$is the trivial bundle, then$R'\to RP^\infty$is the canonical line, so$e(R')=t$and so$w_0(R)=1$and$w_1(R)=0$. You can use this to show that$w_0(V)\in H^0X$is equal to$1$for any bundle over$X$, by pulling back$V$over any point of$X$(where it becomes trivial). If$L\to RP^\infty$is the canonical line, then$L'\to RP^\infty\times RP^\infty$is$L_1\otimes L_2$, the tensor product over of the canonical lines line bundles over each factor. So$e(L')=s+t= 1\cdot t^1 + s\cdot t^0$, giving$w_0(L)=1$and$w_1(L)=s$(where$s\in H^1RP^\infty\$ is the generator).

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