# 18-02-14: Adjoint and Classifying Spaces In general, we define the classifying space $K(G, n)$ (also known as an Eilenberg-MacLane space) to be a space $X$ such that $\pi_n(X) = G$ and for $k\neq n,~\pi_k(X) = 0$. *Note: in my notation, I will simply write this as $\pi_*(X) = G\delta_n$* It is worth mentioning here that there are nice Serre spectral sequences for this family of fibrations: ` \begin{align*} K({\mathbb{Z}}, n-1) \to {\operatorname{pt}}\to K({\mathbb{Z}}, n)\end{align*} `{=html} By examining an appropriate spectral sequence, we were able to find that $H_*({\mathbb{RP}}^\infty) = {\mathbb{Z}}_2\delta_1$, which makes ${\mathbb{RP}}^\infty$ an geometric model of the classifying space $K({\mathbb{Z}}_2, 1)$. Recall that ${\mathbb{CP}}^\infty$ is defined as the limit of the sequence of inclusions ` \begin{align*}{\mathbb{CP}}^1 \subset {\mathbb{CP}}^2 \subset {\mathbb{CP}}^3 \subset \ldots\end{align*} `{=html} together with the weak limit topology. There are a handful of easily recognizable geometric models for a few other types of classifying spaces. $G$ \\ $n$ 1 2 3 ----------------------- ------------------------ ------------------------ ---------------- ${\mathbb{Z}}$ $S^1$ ${\mathbb{CP}}^\infty$ No good model! ${\mathbb{Z}}_2$ ${\mathbb{RP}}^\infty$ $\cdot$ $\cdot$ ${\mathbb{Z}}_p$ $L(\infty, p)$ $\cdot$ $\cdot$ $\ast_n {\mathbb{Z}}$ $\bigvee_n S^1$ $\cdot$ $\cdot$ *Note: $\ast_n {\mathbb{Z}}$ is the free group on $n$ generators. Also, these spaces can all be constructed as a CW complex for any given $G$ - just start with some $\bigvee S^1$ and add cells to kill off all higher homotopy.* Using spectral sequences, we also found that $K({\mathbb{Z}}, 3)$ was a space that, although simple from the point of view of homotopy, had a more complicated structure in homology. It was a number of odd properties- it has torsion in infinitely many dimensions, doesn't satisfy Poincare duality (even in a truncated sense). ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- Consider the fibration ` \begin{align*} S^1 \to S^{2\infty + 1} \to {\mathbb{CP}}^\infty \end{align*} `{=html} where these infinite-dimensional spaces are defined using the weak topology. There is a perfectly good filtration arising from the inclusions in this diagram: ` \begin{align*} \begin{CD} S^3 @>\subseteq>> S^5 @>\subseteq>> S^7 @>\subseteq>> \cdots\\ @VVV @VVV @VVV \\ {\mathbb{CP}}^1 @>\subseteq>> {\mathbb{CP}}^3 @>\subseteq>> {\mathbb{CP}}^5 @>\subseteq>> \cdots \end{CD} \end{align*} `{=html} So we can apply the usual spectral sequence to this filtration. We know that $E_\infty$ can only contain ${\mathbb{Z}}$ in dimension zero, and we obtain the following $E_2$ page: `latex {cmd=true, hide=true, run_on_save=true} \documentclass{standalone} \usepackage{tikz} \usepackage{dsfont} \usepackage{amsmath, amsthm, amssymb} \usetikzlibrary{cd} \begin{document} \begin{tikzcd} \mathbb{Z} \arrow[rrd, "d_2 \cong"] & 0 & 0 \arrow[rrd, "d_2 \cong"] & 0 & 0 \\ 0 & 0 & \mathbb{Z} & 0 & \mathbb{Z} \end{tikzcd} \end{document}` Since $d_2$ is an isomorphism, it must take generators to generators, and so we can deduce the following facts: \- $d_2(\alpha \otimes 1) = 1\otimes\beta$ - $d_2(1\otimes\beta) = 0$ We can now compute ` \begin{align*} d_2(\alpha\otimes\beta) &= d_2(\alpha\otimes 1) \cup (1\otimes\beta) + 0\\ &= 1\otimes\beta^2 \end{align*} `{=html} And using the cup product structure on cohomology, we can fill out the following diagram that summarizes these results: `latex {cmd=true, hide=true, run_on_save=true} \documentclass{standalone} \usepackage{tikz} \usepackage{dsfont} \usepackage{amsmath, amsthm, amssymb} \usetikzlibrary{cd} \newcommand*\ZZ{\mathds{Z}} \begin{document} \begin{tikzcd} \alpha\otimes 1 \arrow[rrd, "d_2 \cong"] & \cdot & \alpha\otimes\beta \arrow[rrd, "d_2 \cong"] & \cdot & \cdot \\ \cdot & \cdot & 1\otimes \beta \arrow[u, "\cup"] & \cdot & 1\otimes\beta^2 \end{tikzcd} \end{document}` Thus, just from knowing that $d_2$ is an isomorphism, we can conclude that $H^4({\mathbb{CP}}^\infty) = {\mathbb{Z}}<\beta^2>$. Alternatively, we'll write this as $H^4({\mathbb{CP}}^\infty) = {\mathbb{Z}}.\beta^2$ By a repeated application of this argument, we find that $H^{2n}({\mathbb{CP}}^\infty) = {\mathbb{Z}}.\beta^n$, allowing us to conclude that ` \begin{align*} H^*({\mathbb{CP}}^\infty) = {\mathbb{Z}}[\beta_{(2)}]. \hfill\blacksquare \end{align*} `{=html} ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- If we know $H^*({\mathbb{CP}}^\infty)$, which is the easiest case, we can then use the inclusion ${\mathbb{CP}}^n \xrightarrow{i} {\mathbb{CP}}^\infty$ (as a cellular map) to induce ` \begin{align*} H^*({\mathbb{CP}}^n) \xrightarrow{i^*} H^*({\mathbb{CP}}^\infty)\\ \beta \mapsto \beta \end{align*} `{=html} which is a actually a *ring* homomorphism instead of just a group homomorphism. This presents a good argument for the use of cohomology, due to its extra ring structure. This is an isomorphism on low-dimensional (co)homology, which reflects the idea encapsulated in the weak limit that these should be approximately equal for large enough $n$. This is indicative of a general principle: if $X$ is a CW complex and $X^n$ is its $n$-skeleton, then the inclusion $X^n \xrightarrow{i} X$ induces an isomorphism $H_k(X^n) \cong H_k(X)$ for $k < n$. (Note that this may or may not be an isomorphism for $k=n$.) In particular, it is again a ring homomorphism, and so carries true relations/equations to true relations/equations. Dually, homology does have *some* type of ring structure, however it is slightly unnatural and onerous to define and use. There is a natural coproduct on $H_*(X)$ for any space $X$, which has a "one in, two out" type and takes this form: ` \begin{align*} H_*(X) \xrightarrow{\Delta} H_*(X) \times H_*(X) \\ a \mapsto \sum a' \otimes a'' \end{align*} `{=html} This coproduct satisfies a form of coassociativity, i.e. if we have ` \begin{align*} \Delta(a) = \sum b_i \otimes c_i \\ (\Delta \otimes 1) \Delta (a) = \sum_{i,j} (d_j^i \otimes e_j^i)\otimes c_i \\ (1 \otimes\Delta ) \Delta (a) = \sum_{i,j} b_i \otimes(f_k^i \otimes g_k^i) \end{align*} `{=html} then the "structure coefficients" agree, i.e. we have $b_i = \sum_j (d_j^i \otimes e_j^i)$ and $c_i = \sum_k (f_k^i \otimes g_k^i)$. In other words, just note that each element on the right hand side of these equations is an element of $H_*^{\otimes 3}$, and so coassociativity simply requires that they are the same element of this space. ------------------------------------------------------------------------ We can specialize by looking at the case where $V$ is a vector space, with a coproduct $V \xrightarrow{\Delta} V\otimes V$. Then pick a basis $\{e_i\}_{i\in I}$, and write ` \begin{align*} \Delta(e_i) = \sum_{j,k} \Delta_i^{j,k}(e_j\otimes e_k) \end{align*} `{=html} where $\Delta_i^{j,k} \in k$, the ground field of $V$. Then coassociativity requires that we have ` \begin{align*} \sum_{j,k,l,m} \Delta_i^{j,k} \Delta_j^{l,m} (e_l \otimes e_m \otimes e_k) = \sum_{j,k}\Delta_i^{j,k} e_j \otimes\Delta_k^{p,q}(e_p \otimes e_q) \end{align*} `{=html} or in other words, that ` \begin{align*} \sum_j \Delta_i^{j,k} \Delta_j^{l,m} = \sum_i \Delta_i^{l,r} \Delta_r^{m,k}\tag*{$\forall k,l,m$} \end{align*} `{=html} It is worth noting that there is also a version of the universal coefficient theorem for homology, which comes in the form ` \begin{align*} 0 \to \operatorname{Ext} (H_{n-1}(X, {\mathbb{Z}}), {\mathbb{Z}}) \to H_n(X, {\mathbb{Z}}) \to \hom(H_n(X,{\mathbb{Z}}),{\mathbb{Z}}) \to 0\end{align*} `{=html} ------------------------------------------------------------------------ One question that comes up here is whether or not there is a sense in which Ext and Hom are "duals" of each other. In some way, this is case, using the "Frobenius duality" of ${-}\otimes R$ and $\hom({-}, S)$. *Aside: Frobenius duality occurs in algebras $A$ over some field $k$ possessing a nondegenerate bilinear form $A\times A \xrightarrow{\sigma} k$ satisfying $\sigma(ab, c) = \sigma(a, bc)$. Such a `\sigma`{=tex}\$ is called a Frobenius norm. A simple example is the trace of a matrix, another example is any Hopf algebra.* This kind of duality comes in the form of something like ` \begin{align*} \hom(M\otimes N, P) = \hom(M, \hom_{\text{in}}(N, P)) \end{align*} `{=html} where $\hom_\text{in}$ is an "internal hom", which is actually an object in the category whose underlying set is the usual $\hom$. One might also call this "map", and denote it $[N, P]$, then the above statement translates to the condition that if $N, P \in \mathcal{C}$ for some category, then $\hom_\text{in}(N, P) = [N, P] \in \mathcal{C}$ is also an object in the same category. (This might also be denoted $\mathcal{Hom}$.) For an analogy, let $\mathcal{C} = \mathbf{Top}$, and $\hom_\mathbf{Top}(X,Y)$ be the set of continuous maps from $X$ to $Y$. Then notice that we can put a topology on this space, say $\mathcal{T}$, so define $\text{Map}(X, Y) = (\hom_\mathbf{Top}(X,Y), \mathcal{T})$, which is in fact an **object** in $\mathbf{Top}$. This becomes the aforementioned "internal hom". Then, the previous adjunction becomes ` \begin{align*} \hom_\textbf{Top}(X\times Y, Z) = \hom_\textbf{Top}(X, \text{Map}(X, Y)) \tag*{$(\in \mathbf{Set})$} \end{align*} `{=html} ------------------------------------------------------------------------ More generally, consider what happens in categories of $R$ modules, where $R$ is generally non-commutative. We can then take objects like $M_R \in \mathbf{mod{\hbox{-}}R}$ and ${}_{R}N_{S} \in \mathbf{R{\hbox{-}}mod{\hbox{-}}S}$. We can then form the tensor product $M_R \otimes_R {}_{R}N_{S}$, and the adjunction becomes ` \begin{align*} \hom_\mathbf{mod{\hbox{-}}S} (M_R \otimes_R {}_{R}N_{S}, P_S) = \hom_\mathbf{mod{\hbox{-}}R} (M_R, \hom_\mathbf{mod{\hbox{-}}S} ({}_{R}N_{S}, P_S)) \tag*{$(\in \mathbf{Ab})$} \end{align*} `{=html} Again, in the second argument of the right-hand side, we identify this as an internal hom - this works because the object $\hom_\mathbf{mod{\hbox{-}}S} ({}_{R}N_{S}, P_S)$ actually becomes a right $R$-module by precomposition. ------------------------------------------------------------------------ In some ways, this resembles the kind of adjunction that occurs in an inner product space - for example, given a matrix $A$, it may have an "adjoint" matrix $A^*$ that satisfies ` \begin{align*} {\left\langle {Av},~{w} \right\rangle} = {\left\langle {w},~{A^*v} \right\rangle} \end{align*} `{=html} and so we can think of $\hom$ like a Hermitian inner product of this form, which is contravariant (re: conjugate) in the first argument. Note that the choice of which argument is contrvariant varies! In Physics, the second argument is often conjugate-linear, while the first is linear. We can also look at this as an almost-commuting of the following diagram \`\`\`latex {cmd=true, hide=true, run_on_save=true}