A note on nontrivial intersection for selfmaps of complex Grassmann manifolds

TL;DR

This paper proves that for certain selfmaps of complex Grassmann manifolds, there always exists a k-plane that intersects its image nontrivially, revealing a fundamental geometric property.

Contribution

It establishes a new intersection property for selfmaps of complex Grassmann manifolds, extending understanding of their geometric structure.

Findings

Existence of a k-plane with nontrivial intersection under any selfmap

Applicable for 1<k<n in complex Grassmann manifolds

Provides insight into the fixed point and intersection theory of these manifolds

Abstract

Let $G(k,n)$ be the complex Grassmann manifold of $k$-planes in $\mathbb C^{k+n}$. In this note, we show that for $1<k<n$ and for any selfmap $f:G(k,n)\to G(k,n)$, there exists a $k$-plane $V^k\in G(k,n)$ such that $f(V^k)\cap V^k\ne \{0\}$.

Equations

$$\mathbb{F}M(n_1,\dots ,n_k)=\frac{U_{\mathbb{F}}(n)}{U_{\mathbb{F}}(n_1)\times \cdots \times U_{\mathbb{F}}(n_k)},$$

$$U_{\mathbb{F}}(n)=\left\{\begin{array}[]{l}O(n)\ \text{the orthogonal group of order}\ n\ \text{if}\ \mathbb{F}=\mathbb{R},\\ U(n)\ \text{the unitary group of order}\ n\ \text{if}\ \mathbb{F}=\mathbb{C},\\ Sp(n)\ \text{the symplectic group of order}\ n\ \text{if}\ \mathbb{F}=\mathbb{H}.\end{array}\right.$$

$$ch({\gamma }^k)=1+c_1+\cdots +c_k,c_i\in H^{2i}(G(k,n);\mathbb{Q}),$$

$$H^{\ast }(G(k,n);\mathbb{Q})=\mathbb{Q}[c_1,\dots ,c_k]/I_{k,n},$$

$${\gamma }^k\oplus {\gamma }^n={ϵ}^{k+n}.$$

$${\bar{c}}_i=\sum _{\Vert \alpha \Vert =i}(-1{)}^{|\alpha |}\frac{|\alpha |!}{\alpha !}ch({\gamma }^k{)}^{\alpha },$$

$$ch({\gamma }^k)⌣ch({\gamma }^n)=ch({ϵ}^{k+n})=1$$

$$(1+c_1+\cdots +c_k)⌣(1+{\bar{c}}_1+\cdots +{\bar{c}}_n)=1$$

$$1$$

$$0$$

$${\bar{c}}_j=-\sum _{i=1}^j{c}_i⌣{\bar{c}}_{j-i}$$

$${\bar{c}}_j=\sum _{||\alpha ||=j}(-1{)}^{|\alpha |}\frac{|\alpha |!}{\alpha !}ch({\gamma }^k{)}^{\alpha },$$

$${\bar{c}}_m$$

$$X(\beta )$$

$$=\sum _{b_i\ne 0}\frac{(|\beta |-1)!b_i}{\beta !}$$

$$=\sum _{i=1}^m\frac{(|\beta |-1)!b_i}{\beta !}$$

$$=\frac{(|\beta |-1)!{\sum }_{i=1}^m{b}_i}{\beta !}$$

$$=\frac{(|\beta |-1)!|\beta |}{\beta !}$$

$$=\frac{|\beta |!}{\beta !}$$

$$ch({\gamma }^k)⌣ch(f^{\ast }{\gamma }^k)⌣ch({\eta }^{n-k})=1.$$

$$ch(f^{\ast }{\gamma }^k)⌣ch({\eta }^{n-k})=1+{\bar{c}}_1+\cdots +{\bar{c}}_n.$$

$$ch(f^{\ast }{\gamma }^k)=1+{\tilde{c}}_1+\cdots +{\tilde{c}}_k,{\tilde{c}}_i\in H^{2i}(G(k,n);\mathbb{Q}),$$

$$ch({\eta }^{n-k})=1+t_1+\cdots +t_{n-k},t_j\in H^{2j}(G(k,n);\mathbb{Q}).$$

$${\bar{c}}_n={\tilde{c}}_k⌣t_{n-k}.$$

$$ch(f^{\ast }{\gamma }^k)=f^{\ast }(ch({\gamma }^k))=1+m{c}_1+m^2{c}_2+\cdots +m^k{c}_k.$$

$${\bar{c}}_n=m^k{c}_k⌣t_{n-k},$$

$$n$$

$$n=(l-1)(k-1)+2(k-2)+3$$

$${\tilde{c}}_4$$

$$t_{3l}$$

$$\alpha {\alpha }^{\prime }$$

$$\alpha \beta$$

$$\theta \beta$$

$$\alpha {\alpha }^{\prime }$$

Full text

1 Introduction

The problem of determining the fixed point property (f.p.p.) for Grassmann manifolds has been studied by many authors (for example [7], [5], [6]).

Let

[TABLE]

$n_{1}+\cdots+n_{k}=n$. Here, $\mathbb{F}$ stands for one of the fields $\mathbb{R}$, $\mathbb{C}$ or the skew field $\mathbb{H}$, and

[TABLE]

In [4], Glover and Homer have given the following necessary condition for $\mathbb{F}M(n_{1},\ldots,n_{k})$ to have the f.p.p..

Theorem 1** ([4], Theorem 1).**

If $\mathbb{F}M(n_{1},\ldots,n_{k})$ has the f.p.p., then $n_{1},\ldots,n_{k}$ are distinct integers and, if $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$, at most one is odd.

The above theorem gives rise to the following conjectures:

Conjecture 1**.**

If $n_{1},\ldots,n_{k}$ are all distinct then $\mathbb{H}M(n_{1},\ldots,n_{k})$ has the f.p.p..

Conjecture 2**.**

If $n_{1},\ldots,n_{k}$ are all distinct and at most one is odd then $\mathbb{F}M(n_{1},\ldots,n_{k})$ has the f.p.p., for $\mathbb{F}=\mathbb{R}$ and $\mathbb{F}=\mathbb{C}$.

The above conjectures were already proved to be true in the following cases:

• •

Projective spaces ($\mathbb{F}M(1,n-1)$);

• •

If $n_{2}$ and $n_{3}$ are distinct positive even integers and $n_{3}\geq 2n_{2}^{2}-1$ then $\mathbb{C}M(1,n_{2},n_{3})$ has the f.p.p. ([4]).

• •

If $1,n_{2}$ and $n_{3}$ are distinct positive integers and $n_{3}\geq 2n_{2}^{2}-1$, then $\mathbb{H}M(1,n_{2},n_{3})$ has the f.p.p. ([4]).

• •

If $n_{2}<n_{3}$ are even integers greater than $1$ and either $n_{2}\leq 6$ or $n_{3}\geq n_{2}^{2}-2n_{2}-2$, then $\mathbb{R}M(1,n_{2},n_{3})$ has the f.p.p. ([4]).

• •

If $n_{1},n_{2},n_{3}$ are positive integers such that at most one is odd, $n_{1}\leq 3$, $n_{3}\geq n_{2}^{2}-1$, and $[n_{1}/2]<[n_{2}/2]<[n_{3}/2]$, then $\mathbb{R}M(n_{1},n_{2},n_{3})$ has the f.p.p. ([4]).

• •

If $\mathbb{F}=\mathbb{C}$ or $\mathbb{H}$, $\mathbb{F}M(2,q)$ has the f.p.p. for all $q>2$ ([7]).

• •

$\mathbb{R}(2,q)$ has the f.p.p. for all $q=4k$ or $q=4k+1$, $k=1,2,3,\ldots$ ([7]).

• •

For $p\leq 3$ and $q>p$ or $p>3$ and $q\geq 2p^{2}-p-1$, $\mathbb{C}M(p,q)$ has the f.p.p. iff $pq$ is even ([5]).

• •

For $p\leq 3$ and $q>p$ or $p>3$ and $q\geq 2p^{2}-p-1$, $\mathbb{H}M(p,q)$ always has the f.p.p. ([5]).

The main tool used to prove the above results is the calculation of the Lefschetz number of a self-map of such a space. Let’s focus on the case of complex Grassmann manifolds $\mathbb{C}(k,n)=G(k,n)$, the space of $k$-planes in $\mathbb{C}^{k+n}$. Let $\gamma^{k}$ be the canonical $k$-plane bundle over $G(k,n)$. If

[TABLE]

is the total Chern class of $\gamma^{k}$, then the cohomology ring $H^{\ast}(G(k,n);\mathbb{Q})$ is given by:

[TABLE]

where $I_{k,n}$ is the ideal generated by the elements $(c^{-1})_{n+1},\ldots,(c^{-1})_{n+k}$. Here, $(c^{-1})_{q}$ is the part of the formal inverse of $c$ in dimension $2q$ (see [6], Theorem 2.1). Then, $c_{1}$ is the only generator in dimension $2$. Therefore, given a self-map $f:G(k,n)\to G(k,n)$, $f^{\ast}(c_{1})=mc_{1}$ for some coefficient $m$.

Theorem 2** ([5], Theorem 1).**

Let $k\leq 3$ and $n>k$ or $k>3$ and $n\geq 2k^{2}-k-1$. Then every graded ring endomorphism of $H^{\ast}(G(k,n);\mathbb{Q})$ is an Adams endomorphism111An Adams endomorphism of $H^{\ast}(G(k,n);\mathbb{Q})$ is a endomorphism $\varphi$ of the form $\varphi(x)=\lambda^{i}x$ for $x\in H^{2i}(G(k,n);\mathbb{Q})$. The coefficient $\lambda$ is called the degree of $\varphi$.. Consequently, if $f:G(k,n)\to G(k,n)$ is a self-map with $f^{\ast}(c_{1})=mc_{1}$ then $f^{\ast}(c_{i})=m^{i}c_{i}$, $i=1,\ldots,k$.

The classification of the graded ring endomorphisms of $H^{\ast}(G(k,n);\mathbb{Q})$ is fundamental in the study of f.p.p. for $G(k,n)$ because of the following.

Proposition 1**.**

An Adams endomorphism of $H^{\ast}(G(k,n);\mathbb{Q})$ has Lefschetz number zero if and only if its degree is $-1$ and $kn$ is odd.

Proof.

See [4], Proposition 4. ∎

In [6], M. Hoffman was able to prove the following.

Theorem 3** ([6], Theorem 1.1).**

Let $k<n$ and $h$ be a graded ring endomorphism of $H^{\ast}(G(k,n);\mathbb{Q})$ with $h(c_{1})=mc_{1}$, $m\neq 0$. Then $h(c_{i})=m^{i}c_{i}$, $1\leq i\leq k$.

If $k<n$ and $h$ is a graded ring endomorphism of $H^{\ast}(G(k,n);\mathbb{Q})$ with $h(c_{1})=0$, it is still unclear about what $h$ looks like in general. The conjecture is that, in this case, $h$ must be the null homomorphism. If one can prove such conjecture then the problem of determining the f.p.p. for $G(k,n)$ will be completely solved.

In this note, we prove a much more modest result for complex Grassmann manifolds than a fixed point theorem. Our main theorem is the following.

Theorem 4** (Main Result).**

Let $k>1$ and $k<n$. Then for every continuous map $f:G(k,n)\to G(k,n)$ there exists a $k$-plane $V^{k}\in G(k,n)$ such that $V^{k}\cap f(V^{k})\neq\{0\}$.

The motivation for this work is the paper [8] where the author gave an alternative proof for the f.p.p. of $\mathbb{C}P^{2n}$ using characteristic classes. In fact, a closer look at the proof of the main result in [8] indicates that the same argument would also yield an alternative proof of the f.p.p. for $\mathbb{R}P^{2n}$ by replacing Chern classes with Stiefel-Whitney classes. We should also point out that a non-trivial intersection result similar to Theorem 4 has been obtained in [1] for maps between two different Grassmann manifolds.

2 Proof of the Main Theorem

Throughout this paper, $G(k,n)$ denotes the complex Grassmann manifold of $k$-planes in $\mathbb{C}^{k+n}$.

Note that, since $G(k,n)$ and $G(n,k)$ are homeomorphic, $\gamma^{k}$ and $\gamma^{n}$ can be seen as subbundles of the trivial bundle $G(k,n)\times\mathbb{C}^{k+n}$, which is denoted by $\epsilon^{k+n}$, and, under such identification,

[TABLE]

Lemma 1**.**

Let $ch(\gamma^{n})=1+\bar{c}_{1}+\cdots+\bar{c}_{n}$ be the total Chern class of the bundle $\gamma^{n}$. Then, a general formula for the class $\bar{c}_{i}$ in terms of the Chern classes of $\gamma^{k}$ is given by

[TABLE]

where $\alpha$ represents the $k$-uple $\alpha=(a_{1},\ldots,a_{k})$, $\|\alpha\|=a_{1}+2a_{2}+\cdots+ka_{k}$, $|\alpha|=a_{1}+a_{2}+\cdots+a_{k}$, $\alpha!=a_{1}!a_{2}!\cdots a_{k}!$ and $ch(\gamma^{k})^{\alpha}=c_{1}^{a_{1}}\smile c_{2}^{a_{2}}\smile\cdots\smile c_{k}^{a_{k}}$.

Proof.

The proof is given recursively in the index $i$.

As $\gamma^{k}\oplus\gamma^{n}=\epsilon^{k+n}$, we have

[TABLE]

in $H^{\ast}(G(k,n);\mathbb{Z})$. So

[TABLE]

and then

[TABLE]

Then

[TABLE]

for all $j=1,\dots,n$, with the convention $c_{i}=0$ when $i>k$. Thus,

(i)

$\bar{c}_{1}=-c_{1}$;

(ii)

$\bar{c}_{2}=-(c_{1}\smile-c_{1})-c_{2}=c_{1}^{2}-c_{2}$;

(iii)

Suppose

[TABLE]

for $j=1,\dots,m-1<n$.

Then

[TABLE]

where

[TABLE]

∎

2.1 Proof of Theorem 4

Suppose, to the contrary, there exists a continuous map $f:G(k,n)\to G(k,n)$ such that $V^{k}\cap f(V^{k})=\{0\}$ for every $k$-plane $V^{k}\in G(k,n)$. Then the direct sum $\gamma^{k}\oplus f^{\ast}\gamma^{k}$ can be seen as a subbundle of the trivial bundle $\epsilon^{k+n}$. Let $\eta^{n-k}$ be the normal bundle of $\gamma^{k}\oplus f^{\ast}\gamma^{k}$ in $\epsilon^{k+n}$. Then

[TABLE]

It follows that

[TABLE]

Let

[TABLE]

and

[TABLE]

We will show that it is impossible for

[TABLE]

The proof of the impossibility of the above equality will be split into several cases.

Case 1: $1<k\leq 3$. Since $c_{1}\in H^{2}(G(k,n);\mathbb{Q})$ is the only generator in dimension $2$, $f^{\ast}(c_{1})$ is a multiple of $c_{1}$, let’s say $f^{\ast}(c_{1})=mc_{1}$. Following [7] and [5], for $k\leq 3$ and $k<n$, every endomorphism of the ring $H^{\ast}(G(k,n);\mathbb{Q})$ that preserves dimension is an Adams endomorphism. Therefore, if $f^{\ast}(c_{1})=mc_{1}$ then $f^{\ast}(c_{2})=m^{2}c_{2},\ldots,f^{\ast}(c_{k})=m^{k}c_{k}$. Thus

[TABLE]

It follows that

[TABLE]

in contradiction with Lemma 1.

Case 2: $k>3$. This case will be split in four cases.

Case 2(i): $n=l(k-1)+r$ with remainder $r\neq 1$, that is, $1<r<k-1$ or $r=0$. In this case, $r$ is of the form $r=2i$ or $r=2i+3$, for some integer $i\geq 0$. In case of $r=2i$, the class $c_{k-1}^{l}\smile c_{2}^{i}$ does not appear in $\tilde{c}_{k}\smile t_{n-k}$ but, by Lemma 1, it appears in $\bar{c}_{n}$, contradicting $\bar{c}_{n}=\tilde{c}_{k}\smile t_{n-k}$. In case of $r=2i+3$, the class $c_{k-1}^{l}c_{2}^{i}c_{3}$ does not appear in $\tilde{c}_{k}\smile t_{n-k}$ but, by Lemma 1, it appears in $\bar{c}_{n}$, contradicting $\bar{c}_{n}=\tilde{c}_{k}\smile t_{n-k}$.

Case 2(ii): $k>4$ and $n=(l+1)(k-1)+1$. In this case, we have

[TABLE]

and, since $n>k$, $l\geq 1$. We can write $n=(l+1)(k-1)+1$ in the form

[TABLE]

and, since we are supposing $k>4$, $k-2>2$. With these information, one can check that the class $c_{k-1}^{m-1}\smile c_{k-2}^{2}\smile c_{3}$ cannot appear in $\tilde{c}_{k}\smile t_{n-k}$. On the other hand, by Lemma 1, the class $c_{k-1}^{m-1}\smile c_{k-2}^{2}\smile c_{3}$ appears in $\bar{c}_{n}$. Therefore, $\bar{c}_{n}=\tilde{c}_{k}\smile t_{n-k}$ is impossible.

Case 2(iii): $k=4$, $n=(l+1)(k-1)+1$ and $l$ even, say $l=2j$. In this case, $n-k=3l$ and, since $n>1$, $l\geq 1$. Let

[TABLE]

Thus, in the product $\tilde{c}_{4}\smile t_{3l}$, $\alpha\alpha^{\prime}$ is the coefficient of $c_{2}^{3j+2}$, $\alpha\beta$ is the coefficient of $c_{2}^{2}\smile c_{3}^{l}$ and $\theta\beta$ is the coefficient of $c_{4}\smile c_{3}^{l}$. From Lemma 1 together with the fact that $\tilde{c}_{4}\smile t_{3l}=\bar{c}_{n}$, it follows that

[TABLE]

Thus

[TABLE]

Then, we conclude that $\alpha=\pm 1$, $\displaystyle\beta=\pm\frac{(l+2)(l+1)}{2}$ and $\displaystyle|\beta|=\frac{(l+2)(l+1)}{2}$ divides $\theta\beta=l+1$. It follows that $l=0$, but $l\geq 1$, a contradiction!

Case 2(iv): $k=4$, $n=(l+1)(k-1)+1$ and $l$ odd, say $l=2j+1$. Again, $n-k=3l$ and, since $n>1$, $l\geq 1$. Let

[TABLE]

It follows that, in the product $\tilde{c}_{4}\smile t_{3l}$, $\alpha\alpha^{\prime}$ is the coefficient of $c_{1}\smile c_{2}^{3j+3}$, $\alpha\beta$ is the coefficient of $c_{2}^{2}\smile c_{3}^{l}$, $\theta\beta$ is the coefficient of $c_{4}\smile c_{3}^{l}$ and $\gamma\beta$ is the coefficient of $c_{1}\smile c_{3}^{l+1}$. Since $\bar{c}_{n}=\tilde{c}_{4}\smile t_{3l}$, together with Lemma 1,

[TABLE]

Thus

[TABLE]

From the two last equalities above, it follows that $\beta$ divides $l+1$ and $l+2$. Therefore, $\beta=1$. It follows that $\alpha=\frac{(l+2)(l+1)}{2}$ and, since $\alpha$ divides $3j+4$,

[TABLE]

Therefore, $l^{2}\leq 3$. Since $l$ is an integer not smaller than $1$, it follows that $l=1$. Then, $3j+4=\frac{3l+5}{2}=4$ is divisible by $\frac{(l+2)(l+1)}{2}=3$, a contradiction! ∎