TestOcr 2009-02-07 00:04:11 2009-02-07 00:04:11 Definition % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here The operation $v/c$ is bilinear, and it is easy to verify that $$ (7.2)\text{ }\quad \delta v/c=v/\partial c+(-1)^{\mathfrak{i}}\delta(v/c). $$ \quad Assume now that $v$ is an equivariant cochain; for ow $\epsilon\pi$ we have $\alpha c=\alpha\Sigma n_{J}e_{f}=\Sigma(\alpha n_{j})(\alpha e_{j})$, then $$ (v/\alpha c)\cdot\sigma=\Sigma(\alpha n_{j})v\cdot(\alpha e_{f})\otimes\sigma=\Sigma(\alpha n_{j})\alpha v\cdot(e_{j}\otimes\sigma) $$ $$ =\alpha^{2}\Sigma n_{j}v\cdot(e_{f}\otimes\sigma)=(v/c)\cdot\sigma. $$ Thus, in this case, \noindent (7.3) $v/\alpha c=v/c$ and $v/(\alpha c-c)=0$. \noindent Consequently, the definition of $v/c$ extends to the case of $v$, an equi- variant cochain, and $c$ an element of $[C_{i}(W;Z_{m}^{\langle q)})]_{\pi}\approx C_{i}(Z_{m}^{(q)}\otimes_{\pi}W)$;the relation (7.2) holds for this extended operation. \quad Now take $v=\emptyset^{\#}u^{n}$ and $c\epsilon C_{i}(Z_{m}^{1q)}\otimes_{\pi}W)$, then $$ \phi\# u^{n}fc\epsilon C^{nq-i}(K;Z_{m}) $$ is defined as the reduction by $c$ of the $n^{\mathrm{t}\mathrm{h}}$ power of $u$. Suppose that $u$ is a cocycle, then $\phi\# u^{n}$ is an equivariant cocycle, and if $c$ is a cycle, it follows from (7.2) that $\phi\# u^{n}/c$ is a cocycle. Moreover, if the cycle $c$ is varied by a boundary, then (7.2) implies that $\phi\# u^{n}/c$ varies by a co- boundary. If $u$ is varied by a coboundary $\phi\# u^{n}/c$ also varies by a coboundary. We only remark here that the proof of this last fact requires a special argument and is not, as in the preceding case, an immediate consequence of (7.2). Thus the class $\{\phi\# u^{n}/c\}$ is a function of the classes $\{u\}, \{c\}$, and it is independent of the particular $\phi_{\#}$, since by (3.1) any two choices of $\phi_{\#}$ are equivariantly homotopic. Then Steenrod defines $\{u\}^{n}/\{c\}$, the reduction by $\{c\}$ of the $n^{\mathrm{t}\mathrm{h}}$ power of $\{u\}$, by $$ \{u\}^{n}/\{c\}=\{\phi\# u^{n}/c\}. $$ This gives the Steenrod reduced power operations; they are operations defined for $u\epsilon H^{q}(K;Z_{m})$ and $c\epsilon H_{i}(\pi;Z_{m}^{\langle q)})$, and the value is $$ u^{n}/c\epsilon H^{nq-i}(K;Z_{m}). $$ \quad In general, the reduced powers $u^{n}/c$ are linear operations in $c$, but may not be linear in $u$. We will list some of their $\mathrm{p}\mathrm{r}\mathrm{o}\varphi$ rties. Unless otherwise stated, we assume $u$ and $c$ as above. \quad First, we have (7.4) $u^{n}/c=0$ if $i>nq-q$. \quad Let $f:K\rightarrow L$ be a map and $f^{*}: H^{q}(L;Z_{m})\rightarrow H^{q}(K;Z_{m})$, the induced homomorphism; then $$ (7.5)\text{ }\quad f^{*}(u^{n}/c)=(f^{*}u)^{n}/c. $$ This result implies topological invariance for reduced powers