S.-T. Yau College Student Mathematics Contests 2022

Problem 1. The topological space $X$$X$XX$X$ is obtained by gluing two tetrahedra as illustrated by the figure. There is a unique way to glue the faces of one tetrahedron to the other so that the arrows are matched. The resulting complex has 2 tetrahedra, 4 triangles, 2 edges and 1 vertex.

Show that $X$$X$XX$X$ can not have the homotopy type of a compact manifold without boundary.

Solution: One can calculate the (Z-coefficient) simplicial homology to see that $H_0(X)=\mathbf{Z},H_1(X)={\mathbf{Z}}^2,H_2(X)=\mathbf{Z}/2$$H_0(X)=\mathbf{Z},H_1(X)={\mathbf{Z}}^2,H_2(X)=\mathbf{Z}/2$H_(0)(X)=Z,H_(1)(X)=Z^(2),H_(2)(X)=Z//2H_{0}(X)=\mathbf{Z}, H_{1}(X)=\mathbf{Z}^{2}, H_{2}(X)=\mathbf{Z} / 2$H_0(X)=\mathbf{Z},H_1(X)={\mathbf{Z}}^2,H_2(X)=\mathbf{Z}/2$, $H_3(X)=\mathbf{Z}$$H_3(X)=\mathbf{Z}$H_(3)(X)=ZH_{3}(X)=\mathbf{Z}$H_3(X)=\mathbf{Z}$. This does not satisfy Poincaré duality, hence the $X$$X$XX$X$ can not have the homotopy type of a compact manifold without boundary. Or one can notice that $X$$X$XX$X$ has Euler characteristic 1, but a closed odd-dimensional manifold has Euler characteristic 0 .

Problem 2. Suppose ( $M,h$$M,h$M,hM, h$M,h$ ) is a closed (i.e., compact without boundary) Riemannian manifold, and $h$$h$hh$h$ is a metric on $M$$M$MM$M$ with $\sec (h)\le -1$$\sec (h)\le -1$sec(h) <= -1\sec (h) \leq-1$\sec (h)\le -1$, where $\sec (h)$$\sec (h)$sec(h)\sec (h)$\sec (h)$ is the sectional curvature. Suppose $\mathrm{\Sigma }$$\mathrm{\Sigma }$Sigma\Sigma$\mathrm{\Sigma }$ is a closed minimal surface with genus $g$$g$gg$g$ in $(M,h)$$(M,h)$(M,h)(M, h)$(M,h)$. Show that

Solution: For any surface $\mathrm{\Sigma }$$\mathrm{\Sigma }$Sigma\Sigma$\mathrm{\Sigma }$ in a Riemannian manifold $(M,h)$$(M,h)$(M,h)(M, h)$(M,h)$, let $x\in \mathrm{\Sigma }$$x\in \mathrm{\Sigma }$x in Sigmax \in \Sigma$x\in \mathrm{\Sigma }$, and $\{e_1,e_2,e_3,e_4,\dots ,e_n\}$$\left\{e_1,e_2,e_3,e_4,\dots ,e_n\right\}${e_(1),e_(2),e_(3),e_(4),dots,e_(n)}\left\{e_{1}, e_{2}, e_{3}, e_{4}, \ldots, e_{n}\right\}$\{e_1,e_2,e_3,e_4,\dots ,e_n\}$ be a local orthonormal frame of $M$$M$MM$M$ at $x$$x$xx$x$ where $e_1$$e_1$e_(1)e_{1}$e_1$ and $e_2$$e_2$e_(2)e_{2}$e_2$ are tangent to $\mathrm{\Sigma }$$\mathrm{\Sigma }$Sigma\Sigma$\mathrm{\Sigma }$ and $e_3,\dots ,e_n$$e_3,\dots ,e_n$e_(3),dots,e_(n)e_{3}, \ldots, e_{n}$e_3,\dots ,e_n$ are normal to $\mathrm{\Sigma }$$\mathrm{\Sigma }$Sigma\Sigma$\mathrm{\Sigma }$. The Gauss equation shows that

Here, $K_{12}$$K_{12}$K_(12)K_{12}$K_{12}$ is the sectional curvature of $T_x\mathrm{\Sigma }\subset TM$$T_x\mathrm{\Sigma }\subset TM$T_(x)Sigma sub TMT_{x} \Sigma \subset T M$T_x\mathrm{\Sigma }\subset TM$. Integrate this identity over $\mathrm{\Sigma }$$\mathrm{\Sigma }$Sigma\Sigma$\mathrm{\Sigma }$ and use the Gauss-Bonnet theorem; we get

Equivalently,

where $g$$g$gg$g$ is the genus of $\mathrm{\Sigma }$$\mathrm{\Sigma }$Sigma\Sigma$\mathrm{\Sigma }$. $\mathrm{\Sigma }$$\mathrm{\Sigma }$Sigma\Sigma$\mathrm{\Sigma }$ being minimal implies that $H=0$$H=0$H=0H=0$H=0$, and $\sec (h)\le -1$$\sec (h)\le -1$sec(h) <= -1\sec (h) \leq-1$\sec (h)\le -1$ implies that $K_{12}\le -1$$K_{12}\le -1$K_(12) <= -1K_{12} \leq-1$K_{12}\le -1$. Se we get

Problem 3. For any topological space $X$$X$XX$X$, the $n$$n$nn$n$-th symmetric product of $X$$X$XX$X$ is the quotient of the Cartesian product $(X{)}^n$$(X{)}^n$(X)^(n)(X)^{n}$(X{)}^n$ by the action of the symmetric group $S_n$$S_n$S_(n)S_{n}$S_n$, which permutes the factors in $(X{)}^n$$(X{)}^n$(X)^(n)(X)^{n}$(X{)}^n$. This space is denoted by ${\mathrm{S}\mathrm{P}}^n(X)$${\mathrm{S}\mathrm{P}}^n(X)$SP^(n)(X)\mathrm{SP}^{n}(X)${\mathrm{S}\mathrm{P}}^n(X)$, and the topology is the natural quotient topology induced from $(X{)}^n$$(X{)}^n$(X)^(n)(X)^{n}$(X{)}^n$.

Show that ${\mathrm{S}\mathrm{P}}^n\left(\mathbf{C}{\mathbf{P}}^1\right)$${\mathrm{S}\mathrm{P}}^n\left(\mathbf{C}{\mathbf{P}}^1\right)$SP^(n)(CP^(1))\mathrm{SP}^{n}\left(\mathbf{C} \mathbf{P}^{1}\right)${\mathrm{S}\mathrm{P}}^n\left(\mathbf{C}{\mathbf{P}}^1\right)$ is homeomorphic to $\mathbf{C}\mathbf{P}{}^n$$\mathbf{C}\mathbf{P}{}^n$CP^(n)\mathbf{C P}{ }^{n}$\mathbf{C}\mathbf{P}{}^n$. Here $\mathbf{C}{\mathbf{P}}^1$$\mathbf{C}{\mathbf{P}}^1$CP^(1)\mathbf{C} \mathbf{P}^{1}$\mathbf{C}{\mathbf{P}}^1$ and $\mathbf{C}\mathbf{P}{}^n$$\mathbf{C}\mathbf{P}{}^n$CP^(n)\mathbf{C P}{ }^{n}$\mathbf{C}\mathbf{P}{}^n$ are equipped with the manifold topology.

Solution: ${\mathbf{C}\mathbf{P}}^n$${\mathbf{C}\mathbf{P}}^n$CP^(n)\mathbf{C P}^{n}${\mathbf{C}\mathbf{P}}^n$ can be interpreted as the space of homogeneous polynomials in two variables of degree $n$$n$nn$n$ modulo multiplication by a non-zero complex constant. Each polynomial is determined up to a constant complex number by its $n$$n$nn$n$ complex roots on $\mathbf{C}\mathbf{P}{\mathbf{P}}^1$$\mathbf{C}\mathbf{P}{\mathbf{P}}^1$CPP^(1)\mathbf{C P} \mathbf{P}^{1}$\mathbf{C}\mathbf{P}{\mathbf{P}}^1$. On the other hand, ${\mathrm{S}\mathrm{P}}^n\left({\mathbf{C}\mathbf{P}}^1\right)$${\mathrm{S}\mathrm{P}}^n\left({\mathbf{C}\mathbf{P}}^1\right)$SP^(n)(CP^(1))\mathrm{SP}^{n}\left(\mathbf{C P}^{1}\right)${\mathrm{S}\mathrm{P}}^n\left({\mathbf{C}\mathbf{P}}^1\right)$ is exactly the $n$$n$nn$n$-tuples of unordered points in $\mathbf{C}\mathbf{P}{}^1$$\mathbf{C}\mathbf{P}{}^1$CP^(1)\mathbf{C P}{ }^{1}$\mathbf{C}\mathbf{P}{}^1$. This induces a bijection $F:{\mathrm{S}\mathrm{P}}^n\left({\mathbf{C}\mathbf{P}}^1\right)\to \mathbf{C}\mathbf{P}{}^n$$F:{\mathrm{S}\mathrm{P}}^n\left({\mathbf{C}\mathbf{P}}^1\right)\to \mathbf{C}\mathbf{P}{}^n$F:SP^(n)(CP^(1))rarrCP^(n)F: \mathrm{SP}^{n}\left(\mathbf{C P}^{1}\right) \rightarrow \mathbf{C P}{ }^{n}$F:{\mathrm{S}\mathrm{P}}^n\left({\mathbf{C}\mathbf{P}}^1\right)\to \mathbf{C}\mathbf{P}{}^n$.

It remains to show $F$$F$FF$F$ and $F^{-1}$$F^{-1}$F^(-1)F^{-1}$F^{-1}$ are both continuous. One direction is relatively easy: because the coefficients of the polynomials are determined by the roots via Vieta's formulas, and Vieta's formulas are polynomials, $F$$F$FF$F$ is continuous. For the other direction, notice that ${\mathrm{S}\mathrm{P}}^n\left({\mathbf{C}\mathbf{P}}^1\right)$${\mathrm{S}\mathrm{P}}^n\left({\mathbf{C}\mathbf{P}}^1\right)$SP^(n)(CP^(1))\mathrm{SP}^{n}\left(\mathbf{C P}^{1}\right)${\mathrm{S}\mathrm{P}}^n\left({\mathbf{C}\mathbf{P}}^1\right)$ is compact (because it is the quotient of a compact space), ${\mathbf{C}\mathbf{P}}^n$${\mathbf{C}\mathbf{P}}^n$CP^(n)\mathbf{C P}^{n}${\mathbf{C}\mathbf{P}}^n$ is Hausdorff, and $F$$F$FF$F$ is a continuous bijection, so $F^{-1}$$F^{-1}$F^(-1)F^{-1}$F^{-1}$ is also a continuous bijection.

Problem 4. Let $M$$M$MM$M$ be a complete noncompact Riemannian manifold. $M$$M$MM$M$ is said to have the geodesic loops to infinity property if for any $[\alpha ]\in {\pi }_1(M)$$[\alpha ]\in {\pi }_1(M)$[alpha]inpi_(1)(M)[\alpha] \in \pi_{1}(M)$[\alpha ]\in {\pi }_1(M)$ and any compact subset $K\subset M$$K\subset M$K sub MK \subset M$K\subset M$, there is a geodesic loop $\beta \subset M\mathrm{\setminus }K$$\beta \subset M\mathrm{\setminus }K$beta sub M\\K\beta \subset M \backslash K$\beta \subset M\mathrm{\setminus }K$, such that $\beta$$\beta$beta\beta$\beta$ is homotopic to $\alpha$$\alpha$alpha\alpha$\alpha$.

Show that if a complete noncompact Riemannian manifold $M$$M$MM$M$ does not have the geodesic loops to infinity property, then there is a line in the universal cover $\tilde{M}$$\tilde{M}$widetilde(M)\widetilde{M}$\tilde{M}$.

Remark: A line is a geodesic $\gamma :(-\mathrm{\infty },\mathrm{\infty })\to M$$\gamma :(-\mathrm{\infty },\mathrm{\infty })\to M$gamma:(-oo,oo)rarr M\gamma:(-\infty, \infty) \rightarrow M$\gamma :(-\mathrm{\infty },\mathrm{\infty })\to M$ such that dist $(\gamma (s),\gamma (t))=|s-t|$$(\gamma (s),\gamma (t))=|s-t|$(gamma(s),gamma(t))=|s-t|(\gamma(s), \gamma(t))=|s-t|$(\gamma (s),\gamma (t))=|s-t|$; a geodesic loop is a curve $\beta :[0,1]\to M$$\beta :[0,1]\to M$beta:[0,1]rarr M\beta:[0,1] \rightarrow M$\beta :[0,1]\to M$ that is a geodesic and $\beta (0)=\beta (1)$$\beta (0)=\beta (1)$beta(0)=beta(1)\beta(0)=\beta(1)$\beta (0)=\beta (1)$.

Solution: Suppose $[\alpha ]\in {\pi }_1(M)$$[\alpha ]\in {\pi }_1(M)$[alpha]inpi_(1)(M)[\alpha] \in \pi_{1}(M)$[\alpha ]\in {\pi }_1(M)$ is a loop that $M$$M$MM$M$ has no geodesic loops to infinity with respect to $\alpha ,K$$\alpha ,K$alpha,K\alpha, K$\alpha ,K$. Suppose $\alpha$$\alpha$alpha\alpha$\alpha$ is based at $x_0$$x_0$x_(0)x_{0}$x_0$. Let $K$$K$KK$K$ be a compact subset $K\subset B_R\left(x_0\right)\subset M$$K\subset B_R\left(x_0\right)\subset M$K subB_(R)(x_(0))sub MK \subset B_{R}\left(x_{0}\right) \subset M$K\subset B_R\left(x_0\right)\subset M$. Let us choose $x_i\in M$$x_i\in M$x_(i)in Mx_{i} \in M$x_i\in M$ with $\mathrm{dist}(x_0,x_i)>R$$\mathrm{dist}\left(x_0,x_i\right)>R$dist(x_(0),x_(i)) > R\operatorname{dist}\left(x_{0}, x_{i}\right)>R$\mathrm{dist}(x_0,x_i)>R$. Minimize curves passing through $x_i$$x_i$x_(i)x_{i}$x_i$ in the homotopy class $[\alpha ]$$[\alpha ]$[alpha][\alpha]$[\alpha ]$ to get a geodesic loop ${\gamma }_i$${\gamma }_i$gamma_(i)\gamma_{i}${\gamma }_i$ that is based at $x_i$$x_i$x_(i)x_{i}$x_i$. Because $M$$M$MM$M$ has no geodesic loops to infinity with respect to $\alpha ,{\gamma }_i$$\alpha ,{\gamma }_i$alpha,gamma_(i)\alpha, \gamma_{i}$\alpha ,{\gamma }_i$ intersects with $K$$K$KK$K$; let $y_i\in K\cap {\gamma }_i$$y_i\in K\cap {\gamma }_i$y_(i)in K nngamma_(i)y_{i} \in K \cap \gamma_{i}$y_i\in K\cap {\gamma }_i$.

Now we go to the universal cover $\tilde{M}$$\tilde{M}$tilde(M)\tilde{M}$\tilde{M}$, and consider the lift ${\tilde{\gamma }}_i$${\tilde{\gamma }}_i$tilde(gamma)_(i)\tilde{\gamma}_{i}${\tilde{\gamma }}_i$ of ${\gamma }_i$${\gamma }_i$gamma_(i)\gamma_{i}${\gamma }_i$, such that ${\tilde{\gamma }}_i$${\tilde{\gamma }}_i$tilde(gamma)_(i)\tilde{\gamma}_{i}${\tilde{\gamma }}_i$ connects ${\tilde{x}}_i$${\tilde{x}}_i$tilde(x)_(i)\tilde{x}_{i}${\tilde{x}}_i$ and $[\alpha ]{\tilde{x}}_i$$[\alpha ]{\tilde{x}}_i$[alpha] tilde(x)_(i)[\alpha] \tilde{x}_{i}$[\alpha ]{\tilde{x}}_i$ in the universal cover. We assume ${\tilde{y}}_i$${\tilde{y}}_i$tilde(y)_(i)\tilde{y}_{i}${\tilde{y}}_i$ is the lift of $y_i$$y_i$y_(i)y_{i}$y_i$ lying on ${\tilde{\gamma }}_i$${\tilde{\gamma }}_i$tilde(gamma)_(i)\tilde{\gamma}_{i}${\tilde{\gamma }}_i$. Let us estimate the distance $d_i$$d_i$d_(i)d_{i}$d_i$ between ${\tilde{y}}_i$${\tilde{y}}_i$tilde(y)_(i)\tilde{y}_{i}${\tilde{y}}_i$ and $[\alpha ]x_i$$[\alpha ]x_i$[alpha]x_(i)[\alpha] x_{i}$[\alpha ]x_i$. Because ${\tilde{\gamma }}_i$${\tilde{\gamma }}_i$tilde(gamma)_(i)\tilde{\gamma}_{i}${\tilde{\gamma }}_i$ is a minimizing geodesic segment, we have $\mathrm{dist}({\tilde{y}}_i,[\alpha ]{\tilde{x}}_i)$$\mathrm{dist}\left({\tilde{y}}_i,[\alpha ]{\tilde{x}}_i\right)$dist( tilde(y)_(i),[alpha] tilde(x)_(i))\operatorname{dist}\left(\tilde{y}_{i},[\alpha] \tilde{x}_{i}\right)$\mathrm{dist}({\tilde{y}}_i,[\alpha ]{\tilde{x}}_i)$ equals to the length of the geodesic line segment ${\tilde{\gamma }}_i$${\tilde{\gamma }}_i$tilde(gamma)_(i)\tilde{\gamma}_{i}${\tilde{\gamma }}_i$ from ${\tilde{y}}_i$${\tilde{y}}_i$tilde(y)_(i)\tilde{y}_{i}${\tilde{y}}_i$ to $[\alpha ]{\tilde{x}}_i$$[\alpha ]{\tilde{x}}_i$[alpha] tilde(x)_(i)[\alpha] \tilde{x}_{i}$[\alpha ]{\tilde{x}}_i$. This is exactly the length of the part of the geodesic loop that connects $y_i$$y_i$y_(i)y_{i}$y_i$ and $x_i$$x_i$x_(i)x_{i}$x_i$. By the triangle inequality, $d_i\ge m_i-R$$d_i\ge m_i-R$d_(i) >= m_(i)-Rd_{i} \geq m_{i}-R$d_i\ge m_i-R$. Similarly, the distance $e_i$$e_i$e_(i)e_{i}$e_i$ between ${\tilde{y}}_i$${\tilde{y}}_i$tilde(y)_(i)\tilde{y}_{i}${\tilde{y}}_i$ and $x_i$$x_i$x_(i)x_{i}$x_i$ satisfies the bound $e_i\ge m_i-R$$e_i\ge m_i-R$e_(i) >= m_(i)-Re_{i} \geq m_{i}-R$e_i\ge m_i-R$.

Therefore, there is a geodesic starting from ${\tilde{y}}_i$${\tilde{y}}_i$tilde(y)_(i)\tilde{y}_{i}${\tilde{y}}_i$ that extends to both directions with length longer than $m_i-R$$m_i-R$m_(i)-Rm_{i}-R$m_i-R$. Notice that $y_i\in K$$y_i\in K$y_(i)in Ky_{i} \in K$y_i\in K$, so for any $i$$i$ii$i$, we can choose ${\tilde{y}}_i$${\tilde{y}}_i$tilde(y)_(i)\tilde{y}_{i}${\tilde{y}}_i$ in some fixed compact domain of $\tilde{M}$$\tilde{M}$widetilde(M)\widetilde{M}$\tilde{M}$. Then as $m_i\to \mathrm{\infty }$$m_i\to \mathrm{\infty }$m_(i)rarr oom_{i} \rightarrow \infty$m_i\to \mathrm{\infty }$, we can pass to a subsequence of ${\tilde{y}}_i$${\tilde{y}}_i$tilde(y)_(i)\tilde{y}_{i}${\tilde{y}}_i$ to get a limit ${\tilde{y}}_{\mathrm{\infty }}$${\tilde{y}}_{\mathrm{\infty }}$tilde(y)_(oo)\tilde{y}_{\infty}${\tilde{y}}_{\mathrm{\infty }}$, and a line passing through this point.

Problem 5. A topological space $X$$X$XX$X$ is called an $H$$H$HH$H$-space if there exist $e\in X$$e\in X$e in Xe \in X$e\in X$ and $\mu :X\times X\to X$$\mu :X\times X\to X$mu:X xx X rarr X\mu: X \times X \rightarrow X$\mu :X\times X\to X$ such that $\mu (e,e)=e$$\mu (e,e)=e$mu(e,e)=e\mu(e, e)=e$\mu (e,e)=e$ and the maps $x\to \mu (e,x)$$x\to \mu (e,x)$x rarr mu(e,x)x \rightarrow \mu(e, x)$x\to \mu (e,x)$ and $x\to \mu (x,e)$$x\to \mu (x,e)$x rarr mu(x,e)x \rightarrow \mu(x, e)$x\to \mu (x,e)$ are both homotopic to the identity map.

(a) Show that the fundamental group of an H -space is Abelian.

(b) Show that the sphere $S^{2022}$$S^{2022}$S^(2022)S^{2022}$S^{2022}$ is not an H-space.

Historic Remark: "H" was suggested by Jean-Pierre Serre in recognition of the contributions in Topology by Heinz Hopf.

(a) Let $[f]$$[f]$[f][f]$[f]$ and $[g]$$[g]$[g][g]$[g]$ be two elements in the fundamental group of $X$$X$XX$X$. We may assume $f:[0,1]\to X$$f:[0,1]\to X$f:[0,1]rarr Xf:[0,1] \rightarrow X$f:[0,1]\to X$ and $g:[0,1]\to X$$g:[0,1]\to X$g:[0,1]rarr Xg:[0,1] \rightarrow X$g:[0,1]\to X$ are both continuous maps with $f(0)=f(1)=g(0)=g(1)=e$$f(0)=f(1)=g(0)=g(1)=e$f(0)=f(1)=g(0)=g(1)=ef(0)=f(1)=g(0)=g(1)=e$f(0)=f(1)=g(0)=g(1)=e$.

Now we define a map $F:[0,1]\times [0,1]\to X$$F:[0,1]\times [0,1]\to X$F:[0,1]xx[0,1]rarr XF:[0,1] \times[0,1] \rightarrow X$F:[0,1]\times [0,1]\to X$ by $F(x,y)=\mu (f(x),g(y))$$F(x,y)=\mu (f(x),g(y))$F(x,y)=mu(f(x),g(y))F(x, y)=\mu(f(x), g(y))$F(x,y)=\mu (f(x),g(y))$. Then $F(\cdot ,0)$$F(\cdot ,0)$F(*,0)F(\cdot, 0)$F(\cdot ,0)$ is homotopic to $f$$f$ff$f$ and $F(0,\cdot )$$F(0,\cdot )$F(0,*)F(0, \cdot)$F(0,\cdot )$ is homotopic to $g$$g$gg$g$. It is clear that

is a homotopy from a curve representing $[f]\cdot [g]$$[f]\cdot [g]$[f]*[g][f] \cdot[g]$[f]\cdot [g]$ to a curve representing $[g]\cdot [f]$$[g]\cdot [f]$[g]*[f][g] \cdot[f]$[g]\cdot [f]$. Therefore $[f]\cdot [g]=[g]\cdot [f]$$[f]\cdot [g]=[g]\cdot [f]$[f]*[g]=[g]*[f][f] \cdot[g]=[g] \cdot[f]$[f]\cdot [g]=[g]\cdot [f]$, and hence ${\pi }_1(X)$${\pi }_1(X)$pi_(1)(X)\pi_{1}(X)${\pi }_1(X)$ is Abelian.

(b) We will show that $S^{2n}$$S^{2n}$S^(2n)S^{2 n}$S^{2n}$ is not a $H$$H$HH$H$-space. In the following we consider $\mathbf{R}$$\mathbf{R}$R\mathbf{R}$\mathbf{R}$-coefficient cohomology. Suppose $S^k$$S^k$S^(k)S^{k}$S^k$ is an H -space, then the map $\mu$$\mu$mu\mu$\mu$ induces

and for a generator $x\in H^k\left(S^k\right),{\mu }^{\ast }(x)=1\otimes x+x\otimes 1$$x\in H^k\left(S^k\right),{\mu }^{\ast }(x)=1\otimes x+x\otimes 1$x inH^(k)(S^(k)),mu^(**)(x)=1ox x+x ox1x \in H^{k}\left(S^{k}\right), \mu^{*}(x)=1 \otimes x+x \otimes 1$x\in H^k\left(S^k\right),{\mu }^{\ast }(x)=1\otimes x+x\otimes 1$ (to see this, one can consider the composition $X\hookrightarrow X\times X\stackrel{\mu }{\to }X$$X\hookrightarrow X\times X\stackrel{\mu }{\to }X$X↪X xx Xrarr"mu"XX \hookrightarrow X \times X \xrightarrow{\mu} X$X\hookrightarrow X\times X\stackrel{\mu }{\to }X$, where the inclusion is $a\to (a,e)$$a\to (a,e)$a rarr(a,e)a \rightarrow(a, e)$a\to (a,e)$ or $a\to (e,a))$$a\to (e,a))$a rarr(e,a))a \rightarrow(e, a))$a\to (e,a))$.

The universality of the cup product gives

The left hand side is clearly 0 , and the right hand side is $(1+(-1{)}^{k^2})x\otimes x$$\left(1+(-1{)}^{k^2}\right)x\otimes x$(1+(-1)^(k^(2)))x ox x\left(1+(-1)^{k^{2}}\right) x \otimes x$(1+(-1{)}^{k^2})x\otimes x$. Here, notice that $(a\otimes b)\cup (c\otimes d)=$$(a\otimes b)\cup (c\otimes d)=$(a ox b)uu(c ox d)=(a \otimes b) \cup(c \otimes d)=$(a\otimes b)\cup (c\otimes d)=$ $(-1{)}^{|\mathrm{deg}(b)||\mathrm{deg}(c)|}(a\cup c)\otimes (b\cup d)$$(-1{)}^{|\mathrm{deg}(b)||\mathrm{deg}(c)|}(a\cup c)\otimes (b\cup d)$(-1)^(|deg(b)||deg(c)|)(a uu c)ox(b uu d)(-1)^{|\operatorname{deg}(b)||\operatorname{deg}(c)|}(a \cup c) \otimes(b \cup d)$(-1{)}^{|\mathrm{deg}(b)||\mathrm{deg}(c)|}(a\cup c)\otimes (b\cup d)$. Thus, $S^k$$S^k$S^(k)S^{k}$S^k$ being an H-space implies that $k$$k$kk$k$ is odd.

Remark: In fact, Adams' Hopf invariant one theorem shows that among all the spheres, only $S^0,S^1,S^3,S^7$$S^0,S^1,S^3,S^7$S^(0),S^(1),S^(3),S^(7)S^{0}, S^{1}, S^{3}, S^{7}$S^0,S^1,S^3,S^7$ are $H$$H$HH$H$ spaces.

Problem 6. A hypersurface $\mathrm{\Sigma }\subset {\mathbf{R}}^{\mathbf{n}\mathbf{+}\mathbf{1}}$$\mathrm{\Sigma }\subset {\mathbf{R}}^{\mathbf{n}\mathbf{+}\mathbf{1}}$Sigma subR^(n+1)\Sigma \subset \mathbf{R}^{\mathbf{n + 1}}$\mathrm{\Sigma }\subset {\mathbf{R}}^{\mathbf{n}\mathbf{+}\mathbf{1}}$ is called a shrinker if it satisfies the equation

Here $H$$H$HH$H$ is the mean curvature, which is $-⟨{\mathrm{tr}}_A,\overrightarrow{n}⟩$$-⟨{\mathrm{tr}}_A,\overrightarrow{n}⟩$-(:tr_(A),( vec(n)):)-\left\langle\operatorname{tr}_{A}, \vec{n}\right\rangle$-⟨{\mathrm{tr}}_A,\overrightarrow{n}⟩$ where $A$$A$AA$A$ is the second fundamental form, $x$$x$xx$x$ is the position vector, and $\overrightarrow{n}$$\overrightarrow{n}$vec(n)\vec{n}$\overrightarrow{n}$ is outer unit normal vector.

(a) Show that $S^n(\sqrt{2n})$$S^n(\sqrt{2n})$S^(n)(sqrt(2n))S^{n}(\sqrt{2 n})$S^n(\sqrt{2n})$, the sphere with radius $\sqrt{2n}$$\sqrt{2n}$sqrt(2n)\sqrt{2 n}$\sqrt{2n}$, is a shrinker.

(b) Show that any compact shrinker without boundary must intersect with $S^n(\sqrt{2n})$$S^n(\sqrt{2n})$S^(n)(sqrt(2n))S^{n}(\sqrt{2 n})$S^n(\sqrt{2n})$.

(a) One can calculate that for $S^n(\sqrt{2n}),A=-\frac{1}{\sqrt{2n}}g\overrightarrow{n}$$S^n(\sqrt{2n}),A=-\frac{1}{\sqrt{2n}}g\overrightarrow{n}$S^(n)(sqrt(2n)),A=-(1)/(sqrt(2n))g vec(n)S^{n}(\sqrt{2 n}), A=-\frac{1}{\sqrt{2 n}} g \vec{n}$S^n(\sqrt{2n}),A=-\frac{1}{\sqrt{2n}}g\overrightarrow{n}$, hence $H=\frac{n}{\sqrt{2n}}$$H=\frac{n}{\sqrt{2n}}$H=(n)/(sqrt(2n))H=\frac{n}{\sqrt{2 n}}$H=\frac{n}{\sqrt{2n}}$. Also, $x=\sqrt{2n}\overrightarrow{n}$$x=\sqrt{2n}\overrightarrow{n}$x=sqrt(2n) vec(n)x=\sqrt{2 n} \vec{n}$x=\sqrt{2n}\overrightarrow{n}$, so $S^n(\sqrt{2n})$$S^n(\sqrt{2n})$S^(n)(sqrt(2n))S^{n}(\sqrt{2 n})$S^n(\sqrt{2n})$ satisfies the shrinker's equation.

(b) Suppose $\mathrm{\Sigma }$$\mathrm{\Sigma }$Sigma\Sigma$\mathrm{\Sigma }$ is a closed shrinker. On any hypersurface, ${\mathrm{\nabla }}_{\mathrm{\Sigma }}x=I$${\mathrm{\nabla }}_{\mathrm{\Sigma }}x=I$grad_(Sigma)x=I\nabla_{\Sigma} x=I${\mathrm{\nabla }}_{\mathrm{\Sigma }}x=I$, where $I$$I$II$I$ is the $(n+1)\times (n+1)$$(n+1)\times (n+1)$(n+1)xx(n+1)(n+1) \times(n+1)$(n+1)\times (n+1)$ matrix that is the identity on $T_x\mathrm{\Sigma }\otimes T_x\mathrm{\Sigma }$$T_x\mathrm{\Sigma }\otimes T_x\mathrm{\Sigma }$T_(x)Sigma oxT_(x)SigmaT_{x} \Sigma \otimes T_{x} \Sigma$T_x\mathrm{\Sigma }\otimes T_x\mathrm{\Sigma }$ and vanishes elsewhere, ${\mathrm{\Delta }}_{\mathrm{\Sigma }}x=-H\overrightarrow{n}$${\mathrm{\Delta }}_{\mathrm{\Sigma }}x=-H\overrightarrow{n}$Delta_(Sigma)x=-H vec(n)\Delta_{\Sigma} x=-H \vec{n}${\mathrm{\Delta }}_{\mathrm{\Sigma }}x=-H\overrightarrow{n}$, so

Consider $x_{\text{max}}$$x_{\text{max }}$x_("max ")x_{\text {max }}$x_{\text{max }}$ such that $|x{|}^2$$|x{|}^2$|x|^(2)|x|^{2}$|x{|}^2$ attains the maximum, and $x_{\text{min}}$$x_{\text{min }}$x_("min ")x_{\text {min }}$x_{\text{min }}$ such that $|x{|}^2$$|x{|}^2$|x|^(2)|x|^{2}$|x{|}^2$ attains the minimum. First let us consider $x_{min}\ne 0$$x_{min}\ne 0$x_(min)!=0x_{\min } \neq 0$x_{min}\ne 0$. Differentiating $|x{|}^2$$|x{|}^2$|x|^(2)|x|^{2}$|x{|}^2$ shows that $x_{max}$$x_{max}$x_(max)x_{\max }$x_{max}$ and $x_{min}$$x_{min}$x_(min)x_{\min }$x_{min}$ are normal to the tangent hyperplane, and $⟨x.,\overrightarrow{n}{⟩}^2=|x.{|}^2$$⟨x.,\overrightarrow{n}{⟩}^2=|x.{|}^2$(:x., vec(n):)^(2)=|x.|^(2)\langle x ., \vec{n}\rangle^{2}=|x .|^{2}$⟨x.,\overrightarrow{n}{⟩}^2=|x.{|}^2$ for $\cdot =max$$\cdot =max$*=max\cdot=\max$\cdot =max$ or $\cdot =min$$\cdot =min$*=min\cdot=\min$\cdot =min$. Then ${\mathrm{\Delta }}_{\mathrm{\Sigma }}|x{|}^2\le 0$${\mathrm{\Delta }}_{\mathrm{\Sigma }}|x{|}^2\le 0$Delta_(Sigma)|x|^(2) <= 0\Delta_{\Sigma}|x|^{2} \leq 0${\mathrm{\Delta }}_{\mathrm{\Sigma }}|x{|}^2\le 0$ at $x_{max}$$x_{max}$x_(max)x_{\max }$x_{max}$, hence $2n-{⟨x_{max},\overrightarrow{n}⟩}^2\le 0$$2n-{⟨x_{max},\overrightarrow{n}⟩}^2\le 0$2n-(:x_(max),( vec(n)):)^(2) <= 02 n-\left\langle x_{\max }, \vec{n}\right\rangle^{2} \leq 0$2n-{⟨x_{max},\overrightarrow{n}⟩}^2\le 0$. This implies that ${\left|x_{max}\right|}^2\ge 2n$${\left|x_{max}\right|}^2\ge 2n$|x_(max)|^(2) >= 2n\left|x_{\max }\right|^{2} \geq 2 n${\left|x_{max}\right|}^2\ge 2n$. Similarly, ${\mathrm{\Delta }}_{\mathrm{\Sigma }}|x{|}^2\ge 0$${\mathrm{\Delta }}_{\mathrm{\Sigma }}|x{|}^2\ge 0$Delta_(Sigma)|x|^(2) >= 0\Delta_{\Sigma}|x|^{2} \geq 0${\mathrm{\Delta }}_{\mathrm{\Sigma }}|x{|}^2\ge 0$ at $x_{min}$$x_{min}$x_(min)x_{\min }$x_{min}$, and ${\left|x_{min}\right|}^2\le 2n$${\left|x_{min}\right|}^2\le 2n$|x_(min)|^(2) <= 2n\left|x_{\min }\right|^{2} \leq 2 n${\left|x_{min}\right|}^2\le 2n$. Finally, if $x_{min}=0$$x_{min}=0$x_(min)=0x_{\min }=0$x_{min}=0$, then it is clear ${\left|x_{min}\right|}^2\le 2n$${\left|x_{min}\right|}^2\le 2n$|x_(min)|^(2) <= 2n\left|x_{\min }\right|^{2} \leq 2 n${\left|x_{min}\right|}^2\le 2n$. Therefore $\mathrm{\Sigma }$$\mathrm{\Sigma }$Sigma\Sigma$\mathrm{\Sigma }$ must intersect $S^n(\sqrt{2n})$$S^n(\sqrt{2n})$S^(n)(sqrt(2n))S^{n}(\sqrt{2 n})$S^n(\sqrt{2n})$.