---- > [!theorem] Theorem. ([[Lefschetz fixed point theorem]]) > Let $M$ be a [[smooth manifold|smooth]] $(\mathbb{Z}\text{-})$ [[compact]] [[(homological) orientation of a manifold|oriented]] [[manifold]] of dimension $d$. Let $f:M \to M$ be a [[smooth maps between manifolds|smooth map]]. Denote by > - $\Delta(M)$ the [[Hausdorff iff diagonal is closed|diagonal]] of $M$, $\Delta(M):=\{ (x, x): x \in M \} \subset M \times M$; > - $\Gamma_{f}=(\id_{M} \times f)\big(\Delta(M)\big)$ the [[graph]] of $f$, $\Gamma_{f}=\{ (x, f(x)) : x \in M\} \subset M \times M$. > > We want to understand the set $\text{Fix}(f)=\{ x \in M: f ( x)= x \}=\Delta(M) \cap \Gamma_{f}$ of fixed points of $f$. The Lefschetz fixed point theorem applies [[Poincare duality]] and in particular the [[intersection product]] to specify how many fixed points (if any) there are:[^1] if $\Delta(M)$ and $\Gamma_{f}$ are [[transverse submanifolds|transverse submanifolds]] of $M \times M$, then $\sum_{x \in \text{Fix}(f)} \text{sgn}(x) = \sum_{k=0}^{n}(-1)^{k} \text{Tr}(f^{*}: H^{k}(M; \mathbb{Q}) \to H^{k}(M; \mathbb{Q})),$ > where $\text{sgn}(x)=\text{sgn}\big(\det(\id- df_{x}) \big) \in \{ \pm 1 \}$[^4] for $df_{x}:T_{x}M \to T_{x}M$ the [[differential of a smooth map between smooth manifolds|differential]] of $f$ and $x \in M$. > [!note] Remark. >- Even if you don't want to think about the *signs* this is still a good result. For example, you might compute the RHS to be $15$ in which case you know that there are at least $15$ fixed points. >- Even if you don't want to think about the *signs* nor *transversality*, this is still a good result. If you compute the RHS to be nonzero knowing nothing about transversality, then you still know a fixed point exists. Indeed, *a priori* there are two possibilities: either the intersection is not transverse (and hence nontrivial, because disjoint $\implies$ tranverse!), in which case you know a fixed point exists, or the intersection is transverse and the theorem applies, in which case you know a fixed point exists (because you computed the RHS to be nonzero). ^note > [!basicexample] > An smooth map $\mathbb{C}P^{2n} \to \mathbb{C}P^{2n}$ has a fixed point. > > Indeed, suppose not. Then $\Gamma_{f}$ is transverse to $\Delta$, and by the [[Lefschetz fixed point theorem]] one has $0=\sum_{k}(-1)^{k} \text{Tr}(f^{*}: H^{k} \to H^{k})$ > > We have computed (three times! [[Gysin sequence|1]], [[the perfect Poincare pairing|2]], [[intersection product|3]]) the [[singular cohomology|cohomology ring]] structure on $\mathbb{C}P^{2n}$, it is $H^{*}(\mathbb{C}P^{n})=\mathbb{Z}[x] / x^{2n+1}$ for $x$ a generator in degree $2$. $f^{*}$ is (iso. to) a map between $\mathbb{Z}$, so given by multiplication by some $\lambda$, $f^{*}(x)=\lambda x$, $f^{*}(x^{i})=\lambda^{i} x^{i}$. So the trace of $f^{*}$ is $\lambda^{i}$ or zero depending on parity. When we take the sum above, the trace is zero for $k$ odd, so we get $1+\lambda^{2}+\dots+\lambda^{2n} \in 2\mathbb{Z}+1$ nonzero, a contradiction. > [!proof]+ Proof. ([[Lefschetz fixed point theorem]]) > Each point in $M$ may be viewed as a $0$-[[singular simplex|singular simplex]], and a finite[^5] set of points in $M$, like $\text{Fix}(f)$, can be viewed as a [[singular homology|singular]] $0$-chain. Write $[\text{Fix}(f)]$ for the homology class of $\text{Fix}(f)$. Have the [[intersection product]] $[\text{Fix}(f)]=[\Gamma_{f} \cap \Delta(M)]=[\Gamma_{f}] \cdot [\Delta(M)]\in H_{0}(M) \xrightarrow[\cong]{\varepsilon} \mathbb{Q}.$ > We have $\sum_{x \in \text{Fix}(f)}\text{sgn}(x)=\varepsilon([\text{Fix}(f)])$, but I guess won't say more about this? > > [[cap product|Cap to cup]] gives the following identity: $\big( D_{M} ^{-1}(a) \smile \varphi \big)[M]=\varphi\big( \overbrace{ [M] \frown D_{M} ^{-1} (a) }^{ D_{M} \circ D_{M} ^{-1}(a) } \big)=\varphi(a) \ \ \ \ (*)$ > which we shall use in the computation to come. > > Unwinding the [[intersection product]] definition (including the "[[intersection product|intersection form]] special case", and recalling the notion of [[diagonal class of a manifold|the diagonal class]] $\delta \in H^{d}(M \times M; \mathbb{Q})$ of $M$, we have: $\begin{align} > \varepsilon([\text{Fix}(f)])&=\varepsilon([\Gamma_{f} ] \cdot [\Delta(M)]) \\ > & \\ > &\cong \big(D_{M \times M} ^{-1}([\Gamma_{f}]) \smile \underbrace{ D_{M \times M} ^{-1}([\Delta(M)]) }_{ \delta } \big) [M \times M] \\ > &= \delta(\overbrace{ [\Gamma_{f}] }^{ (\id ,f)_{*} [M]} ) \ \ (\text{Using }(*)) \\ > &= (\id \times f)^{*} \delta([M]), > \end{align}$ > where the last step I think we used the [[cap product|push-pull]] identity for the [[cap product]]: $\delta \big((\id , f)_{*}[M]\big)= (\id , f)_{*}[M] \frown \delta =(\id \times f)_{*} ([M] \frown (\id , f)^{*} \delta)=(\id , f)_{*}\big( (\id , f) ^{*} \delta [M]\big)$ > $(\id , f):M \to M \times M$ has image $\Gamma_{f}$, factors like $M \xrightarrow{ \Delta} M \times M \xrightarrow{ \id \times f}M \times M$. So pullback is $(\id ,f)^{*}=\Delta^{*} \circ (\id \times f)^{*}$ > > Now use that $\delta= \sum_{i} (-1)^{\text{deg }b_{i}}a_{i} \otimes b_{i}$. $f^{*}b_{i}=\sum_{j}c_{ij} b_{j}$ for some $c_{ij}$ corresponding to [[matrix]] of $f$. So $\begin{align} > ((\id , f)^{*} \delta)[M] &= \left( \ \sum_{i} (-1)^{\text{deg }b_{i}} a_{i} \smile f^{*} b_{i} \ \right) [M] \\ > &= \left( \ \sum_{i} (-1)^{\text{deg }b_{i}} a_{i} \smile \sum_{j}c_{ij} b_{j} \ \right) [M]\\ > &= \left( \ \sum_{ij} (-1) ^{\text{deg }b_{i}} c_{ij} a_{i} \smile b_{j} \ \right) [M] \\ > &= \sum_{ij} (-1)^{\text{deg }b_{i}}c_{ij} \delta_{ij} \text{ (dual bases)} \\ > &= \sum_{k=0}^{n} \sum_{i: \text{deg }b_{i}=k} (-1)^{k} c_{i i \\ > } \\ > &= \sum_{k=0}^{n} (-1)^{k} \text{Tr}(f^{*}: H^{k}(M; \mathbb{Q}) \to H^{k}(M ;\mathbb{Q})) > \end{align}$ > which finishes the proof. ---- #### [^1]: It is possible to develop a version of the theorem with less machinery, which can detect whether or not a fixed point exists but cannot say how many. See the notes for Part II. [^2]: [[the perfect Poincare pairing|Recall]] this is given by [[cup product|cupping]] then evaluating against the [[The Thom Theorem for oriented manifolds|fundamental class]] of $M$, i.e., $\langle \alpha, \beta \rangle=(\alpha \smile \beta)[M]$. Recall that, using [[Poincare duality]], we showed it is [[nondegenerate bilinear form|nondegenerate]] (hence the name) when [[universal coefficients theorem for homology|UCT]] applies, which it does because $\mathbb{Q}$ is a [[field]]. So $\langle a_{i}, b_{j} \rangle=\delta_{ij}$, and note that implicitly $\text{deg }a_{i}=d-\text{deg }b_{i}$ (otherwise they don't pair). [^3]: The set $\{ a_{i} \otimes b_{i} \}$ [[when is an element of a tensor product nonzero?|is then a]] [[basis]] for $H^{*}(M; \mathbb{Q}) \otimes H^{*}(M; \mathbb{Q})$, [[Kunneth Theorem for singular cohomology|which recall]] is the [[graded module|graded vector space]] which in grade $n$ is given by $\bigoplus_{k+ \ell=n}H^{k}(M; \mathbb{Q}) \otimes H^{\ell}(M; \mathbb{Q})\underbrace{ = }_{ \text{Kunneth} }H^{n}(M \times M; \mathbb{Q}).$ [^4]: That this [[determinant]] is nonzero (and that the sign is $\pm 1$) turns out to be a consequence of the transversality assumption. [^5]: Has to be finite by transversality + finiteness. ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` Viewing $\Delta(M)$ as a [[singular homology|singular homology class]] [[(co)homology with coefficients|with rational coefficients]], $[\Delta(M)] \in H_{0}(M \times M; \mathbb{Q})$, recall the [[diagonal class of a manifold|diagonal class]] $\delta=D_{M} ^{-1}([\Delta(M)]) \in H^{d}(M \times M; \mathbb{Q})$, its [[Poincare duality|Poincare dual]]. Recall that $\delta$ may be computed as $\begin{align} \delta&=\sum_{i \in I} (-1)^{|b_{i}| } \ a_{i} \otimes b_{i} \in \underbrace{ H^{d}(M \times M; \mathbb{Q}) }_{ \bigoplus_{k+\ell=d}H^{k}(M; \mathbb{Q}) \otimes H^{\ell}(M; \mathbb{Q}) }, \end{align}$ where $\{ a_{i} \}_{i \in I}$ is a [[basis]] of the [[graded module|graded]] $\mathbb{Q}$-[[vector space]] $H^{*}(M;\mathbb{Q})$ and $\{ b_{i} \}_{i \in I}$ its [[dual basis]] wrt the [[the perfect Poincare pairing|Poincare pairing]][^2] $\langle -,- \rangle:H^{k}(M;\mathbb{Q}) \otimes H^{d-k}(M; \mathbb{Q}) \to \mathbb{Q}$.[^3] [[trace of a matrix]] [[determinant of a matrix]]