Homology of (simplicial) spheres and applications · DL Ferrario's Test Web Page

Homology of (simplicial) spheres and applications

Nullhomotopic odd maps, Borsuk, Lusternik and Schnirelmann

Let $\Delta^n$ the standard simplex of dimension $n$ in $\RR^{n+1}$, which is homeomorphic to a $n$-dimensional disk, and $\Sigma^{n}\subset \Delta^{n+1}$ the simplicial $n$-sphere, i.e. the boundary of $\Delta^{n+1}$ (the union of its $n+2$ $n$-dimensional faces), which is homeomorphic to an $n$-dimensional sphere.

The homeomorphisms can be more explicit: let $B\in \Delta^{n+1}$ be its barycenter, and $\varphi\from \Sigma^n\subset \Delta^{n+1} \to \RR^{n+2}$ the map defined as $\varphi(x) = \dfrac{x-B}{|x-B|}$. The image $\varphi(x)$ has norm $1$, and is in the subspace $\sum_ {j} x_ j = 0$, and thus $\varphi$ gives a homeomorphism $\varphi \from \Sigma^n \approx S^{n} \subset \{ \sum_ j x_ j = 0 \} \subset \RR^{n+2} $.

Let $Y$ be a topological space. A continuous map $f\from S^n \to Y$ is homotopic to a constant map if and only if $f$ is the restriction of a map $\tilde f\from D^{n+1} \to Y$, i.e. if and only if $f$ can be extended to a map $\tilde f \from D^{n+1} \to Y$.

Note that $\Sigma^n$ is the union of the $n+2$ faces $F_ j$ of $\Delta^{n+1}$. Also, the images $M_ j = \varphi F_ j$ are $n+2$ closed subsets of $S^ n$, such that $$S^ n = \bigcup_ {j=0} ^{n+1} M_ j$$ and each $M_ j$ does not contain any antipodal points in $S^n$: for each $j=0,\ldots, n+1$, $x\in M_ j \implies -x \not\in M_ j$.

Definition: A map $f\from S^ n \to S^k$ is odd or antipode-preserving if $f(-x) = - f(x)$ for each $x\in S^ n$.

The following proposition are equivalent.
(1) (Lusternik-Schnirelmann-Borsuk) If $M_ 0$, $\ldots$, $M_ n$ is a closed covering of $S^n$, then there exists $j$ such that $M_ j$ contains two antipodal points $x,-x\in M_ j$.
(2) There are no odd maps $f \from S^n \to S^{n-1}$.
(3) (Borsuk's antipodal theorem) Any odd map $f\from S^{n-1} \to S^{n-1}$ is not null-homotopic (homotopic to a constant).
(3') Any map $f\from S^{n-1} \to S^{n-1}$ such that $f(x) \neq f(-x)$ is not null-homotopic (homotopic to a constant).
(4) (Borsuk-Ulam) For each map $f \from S^n \to \RR^n$, there are two antipodal points $x,-x$ in $S^n$ such that $f(x) = f(-x)$.


(1) $\implies$ (2): If $f\from S^n \to S^{n-1}$ is odd, then let $\varphi F_ 0$, $\ldots$, $\varphi F_ n$ be the images of the $n+1$ faces $F_ j$ of $\Delta^n$ in $S^{n-1}$, as above with the origin in the barycenter. The preimages $f^{-1} M_ j$, with $j=0,\ldots, n$, do not contain pairs of antipodal points in $S^n$, and are a closed covering of $S^ n$.

(2) $\iff$ (3): If $f\from S^{n-1} \to S^{n-1}$ is odd and it is homotopic to a constant, then let $D^ n_ +$ and $D^ n_ - = - D^ n_ +$ denote the two hemispheres of $S^{n}$. It is therefore possible to define an odd map $f\from S^n \to S^{n-1}$. Conversely, an odd map $f\from S^n \to S^{n-1}$ yield, by restriction to the equator, an odd map $S^{n-1} \to S^{n-1}$ which is extendable to an hemisphere, and hence nullhomotopic.

(3) $\iff$ (3'): If (3') holds, and $f\from S^{n-1} \to S^{n-1}$ is an odd map, then $f(x) = - f(-x) \neq f(-x)$, and hence $f$ is not null-homotopic by (3'). Conversely, if (3) holds, and $f\from S^{n-1} \to S^{n-1}$ is a null-homotopic map such that $f(x) \neq f(-x)$, then the map $$F(x) = \dfrac{f(x) - f(-x)}{|f(x) - f(-x)|}$$ is odd. Both $f(x)$ and $F(x)$ belong to $S^{n-1}$, and hence the scalar product $f(x) \cdot f(-x) \leq 1$ . This implies that the scalar product $F(x) \cdot f(x)$ satisfies $$ f(x) \cdot F(x) = \dfrac{f(x)\cdot f(x) - f(-x) \cdot f(x)}{|f(x) - f(-x)| } > 0, $$ and hence for each $t\in [0,1]$, $(1-t)f(x) + t F(x) \neq 0$. In fact, otherwise if $(1-t)f(x) + t F(x) = 0$ then $0 = (1-t)f(x)\cdot f(x) +t F(x) \cdot f(x) = (1-t) + t F(x) \cdot f(x) > 0$. since $(1-t)f(x) + t F(x) \neq 0$ for every $x$, the map $$ \dfrac{(1-t)f(x) + t F(x)}{% (1-t)f(x) + t F(x) } $$ is a well-defined homotopy $h_t$, such that $h_0 =f$ and $h_1=F$. Since $f$ is null-homotopic, also $F$ is null-homotopic.

(2) $\implies$ (4): Let $f\from S^n \to \RR^n$ be the map of (4), such that $f(x) \neq f(-x)$ for each $x$. Then it is possible to define $ F\from S^{n} \to S^{n-1}$ defined as $F(x) = \dfrac{f(x) - f(-x)}{|f(x) - f(-x)|}$, which is odd.

(4) $\implies$ (1): Let $M_ 0$, $\ldots$, $M_ n$ be $n+1$ closed subsets of $S^n$ such that $S^n = \bigcup_ {j=0} ^n M_ j$. Assume that $j=0,\ldots, n$ the subset $_ j$ does not contain any pair of antipodal points. Then $M_ j$ and $-M_ j$ are disjoint closed subsets of $S^n$. Let $g_ j\from S^ n \to \RR$ be a continuous function (Urysohn function) such that $g_ j(M_ j) = 0$ and $g_ j(-M_ j) = 1$. If $g_ j(x) = g_ j(-x)$, then $x \not\in M_ j$, and $-x\not\in M_ j$. Furthermore, let $f\from S^ n \to \RR^n$ be the map defined as $$ f(x) = ( g_ 1(x) , \ldots, g_ n(x) ) \in \RR^n . $$ By (4), there exists $x\in S^n$ such that $f(x) = f(-x)$, and therefore there exists $x\in S^n$ such that $g_ j(x) = g_ j(-x)$ for each $j=1,\ldots, n$, which implies that $x\not\in M_ j$ for $j=1,\ldots, n$, and $-x\not\in M_ j$ for $j=1,\ldots, n$.
It must therefore be true that $\pm x\in M_ 0 $, and therefore $M_ 0$ contains the two antipodal points $\pm x$.

Lusternik-Schnirelmann category


If a topological space $X$ is covered by $n+1$ closed subsets $F_ j \subset X$, $j=0\ldots n$, each contractible in $X$ (i.e. such that for each $j$ the inclusion $F_ j \to X$ is homotopic to a constant), then $\Cat X \leq n$. We define the Lusternik-Schnirelmann Category of $X$ $\Cat X$ as the minimum number $n$ such that $X$ is covered by $n+1$ closed subsets, contractible in $X$.

Alternative definitions: As in Fox 1941, a subset $A\subset X$ is categorical if there is an open set $U$ in $X$ which contains $A$ and is contractible in $X$. Then the LS category can be defined as the minimum number of categorical subsets of $X$ that can cover $X$. Or, as in Cornea et al. 2003, it can be defined as the minimum number of open subsets of $X$ which are contractible in $X$ (and thus they are categorical in Fox sense) that can cover $X$. Fox (1941) addresses this problem in §14, for ANRs, but the idea can be traced back to Satz 3 of Borsuk (1936). This is the main point: if $X$ is an ANR (e.g., a manifold) and $A\subset X$ is a closed subset, then $A$ it is contractible in $X$ if and only if $A$ is categorical in $X$. In fact, if $h\from A\times I \to X$ is an homotopy between the inclusion $A\subset X$ and a constant map on $p\in X$, i.e. such that $h(a,0) = p$ and $h(a,1)=a$ for all $a\in A$, then the map $\tilde h$ defined on the closed subspace $Q=X\times \{0\} \cup A \times I \cup X \times \{1\} \subset X\times I$ has a continuous extension to an open neighborhood of $Q$ in $X\times I$, and hence to $U\times I$ for a suitable open $U$ containing $A$.

Recall that if $X$ is compact and Hausdorff, if $A$ is closed in $X$, and if $A$ is contained in $U$ open and contractible in $X$, then $A$ is contained in some $W$ open such that $\overline W \subset U$, where $\overline W$ is the closure of $W$ in $X$.

The following proposition is equivalent to any of the propositions in (3), where $\PP^n(\RR)$ denotes the real projective space of dimension $n$: $$\Cat \PP^n(\RR) = n$$

Proof: Consider the standard projection $p\from S^n \to \PP^n(\RR)$, which sends $\pm x$ to the projective class $[x]$. If $F \subset \PP^n$ is a closed subset, contractible in $\PP^n$, then the inclusion $\iota \from F \to \PP^n$ has a lifting $\iota \from F \to S^n$ (actually, both $\pm\iota$ are liftings). Being a lifting, $p$ is injective when restricted to $\iota F$, and hence $\iota F$ does not contain a pair of antipodal points. For any such closed $F$, there exists $U$ open contractible containing $F$, and $W$ open such that $F\subset W \subset \overline W \subset U$.

Diagram of the projection $p\from S^1 \to \PP^1(\RR)$

Now, if $\PP^n$ is the union of $l$ closed subsets $F_ 1, \ldots , F_ l$, all of which are contractible in $\PP^n$, then by replacing them with the closures $\overline{W}$ of some open subsets $W$ as above, it is possible to assume that $\PP^n$ is the union of the interiors $W_ 1$, $\ldots$, $W_ l$ of the $l$ closed subsets $F_ 1,\ldots, F_ l$, which are contractible in $X$.

In $S^ n$ consider the union $L\subset S^n$ of the $l$ open subsets $V_ 1 = \iota W_ 1$, $\ldots$, $V_ l = \iota W_ l$, none of which contains a pair of antipodal points. The union $L$ need not be equal to $S^n$, but given any pair of antipodal points $\pm x \in S^n$, at least one of them is in $L$. Now, this implies that no pair of antipodal points in the complement $M_ 0 = {S^n \smallsetminus L}$. The same hold for all the corresponding $M_ 1= \iota F_ 1$, $\ldots$, $M_ l =\iota F_ l$. We have thus a cover of $l+1$ closed sets not containing antipodal points of $S^ n$, since $S^ n = M_ 0 \cup L = M_ 0 \cup V_ 1 \cup \ldots \cup V_ l$.

On the other hand, if $M\subset S^n$ is a closed subset which does not contain a pair of antipodal points, then in particular the inclusion $i \from M \to S^n$ is not surjective, and hence it is null-homotopic, as the composition $pi\from M \to \PP^n$. Therefore the image $F=pM\subset \PP^n$ is a closed subset, and since the restriction of $p$ to $M$ is injective, the subspace $F=pM$ is contractible in $\PP^n$. Hence any cover $M_ 0$, $\ldots$, $M_ l$ of closed subsets of $S^n$ not containing antipodal pairs induces a cover $F_ 1$, $\ldots$, $F_ l$ of closed subsets of $\PP^n$ which are contractible in $\PP^n$.

Combining the two arguments, we have proven that proposition (1) of (2) is equivalent to $\Cat \PP^n \geq n$.

Since we have seen above that there are $S^n$ can be covered by $n+2$ contractible subsets without antipodal pairs (the faces of the boundary of $\Delta^{n+1}$), or, equivalently, $\PP^n(\RR)$ is covered by $n+1$ affine charts (which can be refined to a closed cover of contractible spaces), $\Cat \PP^n(\RR) \leq n$, and so we can conclude trhat that proposition (1) of (2) is equivalent to $\Cat \PP^n = n$. /qed/

Lefschetz number, Brouwer and Fixed Points


Definition Let $X=|K|$ be a polyhedron, and $f\from X \to X$ a continuous self-map. For each $j$, let $\Tr H_ j(f)$ denote the trace of the square matrix $$H_ j(f) \from H_ j(X;\QQ) \to H_ j (X; \QQ).$$ The Lefschetz number of $f$ is $$ L(f) = \sum_ {j=0} ^n (-1)^j \Tr H_ j(f). $$

By definition it is a homotopy invariant: if $f_ 0 \sim f_ 1$, then $L(f_ 0) = L (f_ 1)$.

Lefschetz Fixed Point Theorem: If for all $x\in X$, $f(x) \neq x$, then $L(f) = 0$.

Recall that $H_ n(S^n) = \ZZ$. The degree of a self-map $f\from S^n \to S^n$ is the (unique) integer $d\in \ZZ$ such that $H_ n(f)\from H_ n(S^n) \to H_ n(S^n)$ sends $k$ to $dk$. Denote it by $\operatorname{gr}(f)$ or $\deg(f)$.

For every map $f\from S^n \to S^n$, $L(f) = 1 + (-1)^n \deg(f)$.

Proof: It follows from the definition, and a change of integer-rational coefficients. /qed/

Hence one other proposition equivalent to any of (2) is:

Any odd map $f \from S^{n-1} \to S^{n-1}$ has $\deg(f) \neq 0$.

Brief sketch of the proof: Instead of an ordinary simplicial approximation of $f$, one can build a symmetric (with respect to the antipodal map $\pm$) simplicial approximation, and using Hopf trick on the Chain Complex (as with the Euler-Poincaré characteristic), obersve that all $C_ n(f)$ have even traces, and therefore that their alternating sum, which is, $L(f)$ is even. Or, alternatively, note that $L(f)$ is also the sum of all fixed point indices of all (isolated) fixed points of $f$, which are equivariant with respect to the action of the group $\pm 1$. Being $L(f)$ even, $\deg(f)$ cannot be zero.

The identity map $1\from S^n \to S^n$ is not null-homotopic.

Proof: Either with degree as above. Or, as a consequence of (2), since the identity is an antipodal-preserving map. Or, also, as follows: it is null-homotopic if and only if it can be extended to the disk $D^{n+1} \subset S^n \to S^n$. Hence it is null-homotopic if and only if there is a retraction of the disk onto its boundary, i.e. if there exists a map $r\from D^{n+1} \to S^{n}$ such that $ri=1_ {S^n}$, where $i\from S^n \to D^{n+1}$ denotes the inclusion. By functoriality of $H_ n$, since $H_ n(D^{n+1})=0$, if such $r$ exists then $\deg(1_ S^{n})$ cannot be $1$ as it should be by functoriality. /qed/

It is not difficult to prove the following equivalences.

The following propositions are equivalent:
(1) $\Cat S^n =1$.
(2) $S^n$ is not contractible.
(3) (Brouwer) Every map $f\from D^{n+1} \to D^{n+1}$ has at least one fixed point.
(4) There is no retraction $r\from D^{n+1} \to S^{n}$, i.e. there is no continuous $r\from D^{n+1} \to S^n$ such that $r(x) = x$ for each $x\in S^{n}$.

The Brouwer Fixed Point Theorem follows from the Lefschetz Theorem, since $L(f) = 1$ for any map on $D^{n+1}$, and therefore it cannot be zero (as for a fixed-point-free map).

Yang-Krasnosel'skii genus: Let $X\subset \RR^d\smallsetminus 0 $ be a antipodally-symmetric space (i.e. $-X = X$, such as a sphere centered in the origin). The genus $\gamma(X)$ of $X$ is the smallest positive integer $n$ such that there exists an odd map (equivariant map) $f\from X \to S^n$, such that $f(-x) = - f(x)$. Thus, (2) is equivalent to $$ \gamma(S^n) \geq n. $$ Actually, in 1955 Bourgin and Yang proved the following extension of the Borsuk-Ulam theorem (for more general spaces $X$ with a fixed-point-free involution): if $X$ has genus $\gamma(X) \geq n$ and $f\from X \to \RR^k$ is any map, then the set $A_ f = \{ x\in X : f(x) = f(-x) \}$ is closed and antipode-invariant, $\gamma (A_ f) \geq n-k$, and $\dim (A_ f)\geq n-k$.