Smooth Manifolds with Infinite Fundamental Group Admitting No Real Projective Structure

Abstract

It is an important question whether it is possible to put a geometry on a given manifold or not. It is well known that any simply connected closed manifold admitting a real projective structure must be a sphere. Therefore, any simply connected manifold   M   which is not a sphere   $(\dim M \ge 4)$   does not admit a real projective structure. Cooper and Goldman gave an example of a 3-dimensional manifold not admitting a real projective structure and this is the first known example. In this article, by generalizing their work, we construct a manifold   $M^n$   with the infinite fundamental group   ${\mathbb {Z}}_2 *{\mathbb {Z}}_2$,   for any   $n\ge 4$,   admitting no real projective structure.

Appendices

Appendix A

Choosing an Appropriate P Depending on the Matrix A

In this section, we continue choosing an appropriate   P   for   A   and calculate trace(Q) to say that the determinant of the Jacobian matrix at some points is nonzero by considering the following composition:

$$\begin{aligned} {\mathbb {R}}^2 \longrightarrow \mathrm{{GL}}(n+1, {\mathbb {R}}) \longrightarrow \mathrm{{SL}}(n+1, {\mathbb {R}}) \longrightarrow {\mathbb {R}}^2 \end{aligned}$$

given by:

$$\begin{aligned} (x, y) \longmapsto P \longmapsto f(P)=A P A P^{-1} \longmapsto g(Q)= (\text {trace}(Q), \text {trace}(Q^2)). \end{aligned}$$

Case 2: A   has two   $-1$   eigenvalues. Then, we choose   P   as follows:

$\bullet$ If   t   is odd,

set   $k=(t-1)/2$   and   $a_{k1}=y$.   If   $k\ne (t-1)/2$,   let:

$$\begin{aligned} a_{k1} = \left\{ \begin{array}{l@{\quad }l} 1, &{} k\text { is odd},\\ 0, &{} k\text { is even}, \end{array} \right. \end{aligned}$$

$a_{t2}=x,\,\, a_{t(t-1)}=y-x$,   $a_{tk}=0$,   for $3\le k \le t-2$,   $a_{12}=y+x,\,\, a_{1(t-1)}=y$.   If   $t\ne 5$,   take   $a_{1((t+3)/2)}= a_{1((t-1)/2)}=1$;   otherwise,   $a_{1k}=0$,   for   $3\le k \le t-2$   and if   $t=5$,   then   $a_{13}=1.$   When   $k=((t+1)/2)+1$   let   $a_{kt}=x$.   Otherwise, (i.e., $k\ne ((t+1)/2)+1$):

$$\begin{aligned} a_{kt} = \left\{ \begin{array}{l@{\quad }l} 0, &{} k\text { is odd},\\ 1, &{} k\text { is even}, \end{array} \right. \end{aligned}$$

and the core matrix   $(t-2)\times (t-2)$   is the identity matrix.

If   A   has two   $-1$   eigenvalues and   $t= 9$,   then we choose   $P_{t\times t}$   as below:

$$\begin{aligned} \begin{bmatrix} 1 &{}\quad y+x &{}\quad 0 &{}\quad 1 &{}\quad 0 &{}\quad 1 &{}\quad 0 &{}\quad y &{}\quad 0 \\ 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 1\\ 1 &{}\quad 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0\\ y &{}\quad 0 &{}\quad 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 1 \\ 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad x\\ 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 1 &{}\quad 1\\ 1 &{}\quad x &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad y-x &{}\quad 0 \end{bmatrix}. \end{aligned}$$

If $(t-1)/2$ is even, then:

$$\begin{aligned} \begin{aligned} \text {trace}(Q)&= t-6-\frac{2y}{1+y+2x+y^2}-\frac{-y-x}{1+y+2x+y^2}-\frac{1+y^2}{1+y+2x+y^2}\\&\quad -\frac{-1-2x-y^2+yx}{1+y+2x+y^2} -\frac{y(1+x)}{1+y+2x+y^2}-\frac{-y^2}{1+y+2x+y^2}+\frac{y+2x}{1+y+2x+y^2}\\&\quad +\frac{1}{1+y+2x+y^2} +\frac{1+y+x+y^2}{1+y+2x+y^2}+\frac{1+y+x+yx}{1+y+2x+y^2}-\frac{-x-y^2+yx}{1+y+2x+y^2}\\&\quad -\frac{1+2y+2x}{1+y+2x+y^2} -\frac{x(-1+y)}{1+y+2x+y^2}+\frac{(y-x)(-1+y)}{1+y+2x+y^2}, \end{aligned} \end{aligned}$$

where $Q= A P A P^{-1}$.

Considering the same map with the case   t   is even, we get the determinant of the Jacobian matrix at   (2, 3)   is   $\displaystyle -\frac{1792}{4913}$.

If   $(t-1)/2$   is odd, then:

$$\begin{aligned} \begin{aligned} \text {trace}(Q)&=t-6-\frac{3y}{y^2+2y+2x}-\frac{-y-x}{y^2+2y+2x}-\frac{2y^2}{y^2+2y+2x}-\frac{y+x}{y^2+2y+2x}\\&\quad -\frac{-y^2+yx-y-x}{y^2+2y+2x} -\frac{xy+y+x}{y^2+2y+2x}+\frac{y^2+y+x}{y^2+2y+2x}+\frac{x}{y^2+2y+2x}\\&\quad + \frac{yx+2x+y}{y^2+2y+2x} -\frac{y(x-y-1)}{y^2+2y+2x} -\frac{xy}{y^2+2y+2x}+\frac{y(y-x)}{y^2+2y+2x}, \end{aligned} \end{aligned}$$

and the determinant of the Jacobian matrix at   (2, 3)   is   $\displaystyle -\frac{768}{6859}$.

In each case, the determinant of the Jacobian is nonzero, and thus, the image of the map   $f \circ g$   contains an open set.

Case 3: If   A   has more than two   $-1$   eigenvalues, we take   P   as below.

First, consider the following composition:

$$\begin{aligned} {\mathbb {R}}^k \longrightarrow \mathrm{{GL}}(n+1, {\mathbb {R}}) \longrightarrow \mathrm{{SL}}(n+1, {\mathbb {R}}) \longrightarrow {\mathbb {R}}^k, \end{aligned}$$

given by:

$$\begin{aligned} (x_1, x_2, ..., x_k) \longmapsto P \longmapsto f(P)=A P A P^{-1}= Q \longmapsto g(Q), \end{aligned}$$

where   $g(Q)= (\text {trace}(Q), \text {trace} (Q^2), ... , \text {trace}(Q^k))$   and   k   is the number of   $-1$   eigenvalues of   A.   The Jacobian matrix is given by:

$$\begin{aligned} {\mathbf {J}} = \left[ \begin{array}{cccc} \displaystyle \frac{\partial \ \text {trace}(Q)}{\partial x_1} &{} \displaystyle \frac{\partial \ \text {trace}(Q)}{\partial x_2} &{} \dots &{} \displaystyle \frac{\partial \ \text {trace}(Q)}{\partial x_k} \\ \displaystyle \frac{\partial \ \text {trace}(Q^2)}{\partial x_1} &{} \displaystyle \frac{\partial \ \text {trace}(Q^2)}{\partial x_2} &{} \dots &{} \displaystyle \frac{\partial \ \text {trace}(Q^2)}{\partial x_k} \\ \vdots &{} \vdots &{} \vdots &{} \vdots \\ \displaystyle \frac{\partial \ \text {trace}(Q^k)}{\partial x_1} &{} \displaystyle \frac{\partial \ \text {trace}(Q^k)}{\partial x_2} &{} \dots &{} \displaystyle \frac{\partial \ \text {trace}(Q^k)}{\partial x_k} \end{array} \right] . \end{aligned}$$

$\bullet$ If   t   is even,

let $a_{12}=x_2,\quad a_{1(t/2)}=a_{1(t+2)/2}=x_3,\quad a_{1(t-1)}=x_1,\quad a_{2(t-2)}=x_3,$ $a_{(t/2)1}=x_2,\quad a_{((t+2)/2)1}=x_3,\quad a_{(t/2)t}=x_3,\quad a_{((t+2)/2)t}=x_1, a_{(t-1)1}=1,\quad a_{t2}=x_3,\quad a_{t(t-1)}=x_2$,   and all the diagonal elements are   1.

According to the number of   $-1$   eigenvalues of   A,   we determine the number of different variables   $x_i \in {\mathbb {R}}$,   where   $3\le i \le k$   and   $k=t/2$. In the core matrix, on the antidiagonal, there are only   $x_i$’s   (except $x_3$) as a pair, which are symmetric with respect to the diagonal. Moreover, the number of some   $x_i$s   is more than two conforming to the dimension. In addition, other entries of   P   are all   0.

For example, if   A   has six   $-1$   eigenvalues and   $t=14$   then   P   is as below:

$$\begin{aligned} {P} = \left[ \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} 1 &{} x_2 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_3 &{} x_3 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_1 &{} 0 \\ 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_3&{} x_2 &{} 0 \\ 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_2 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_5 &{} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_6 &{} 0 &{} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 &{} 0 &{}0 &{} 1 &{} 0 &{} 0 &{} x_4 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0\\ x_2 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} x_1&{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_3\\ x_3 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_1 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_1\\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_4 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} x_6 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 &{} x_5 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} x_2 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 \\ 1 &{} x_2 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 \\ 0 &{} x_3 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_2 &{} 1 \end{array} \right] . \end{aligned}$$

At the point   (2, 3, 4, 5, 6, 7)   the determinant of the Jacobian is:

$$\begin{aligned} \frac{3203652023}{129225403018523774123952000}. \end{aligned}$$

$\bullet$ If   t   is odd,

let $a_{12}=x_2,\quad a_{1(t+1)/2}=x_3,\quad a_{1(t-1)}=x_1,\quad a_{2(t-2)}=x_3$, $a_{((t-1)/2)1}=x_2,\quad a_{((t+1)/2)1}=1,\quad a_{((t+3)/2)1}=x_3,\quad a_{(t-1)1}=1$, $a_{((t-1)/2)t}=x_3,\quad a_{((t+3)/2)t}=x_1,\quad a_{t2}=x_3,\quad a_{t(t-1)}=x_2$   and the diagonal elements are all   1.

In the core matrix, on the antidiagonal, there are only   $x_i$’s   (except $x_3$) as a pair, which are symmetric with respect to the diagonal. Moreover, the number of some   $x_i$’s   are more than two conforming to the dimension. In addition, other entries of   P   are all   0.

For example, if   A   has five   $-1$   eigenvalues and   $t=13$   then   P   is as follows:

$$\begin{aligned} {P} = \left[ \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} 1 &{} x_2 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_3 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_1 &{} 0 \\ 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_3&{} x_2 &{} 0 \\ 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_2 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_5 &{} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} x_4 &{} 0 &{} 0 &{} 0 &{} 0\\ x_2 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0&{} x_1 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_3\\ 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0\\ x_3 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_1 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_1\\ 0 &{} 0 &{} 0 &{} 0 &{} x_4 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} x_5 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} x_2 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 \\ 1 &{} x_2 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 \\ 0 &{} x_3 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} x_2 &{} 1 \end{array} \right] . \end{aligned}$$

At the point   (2, 3, 4, 5, 6)   the determinant of the Jacobian is:

$$\begin{aligned} \frac{74929536}{42961619719375}. \end{aligned}$$

Case 4: If   A   has eigenvalues   $\pm i$   then both   $+i$   eigenspace and   $-i$   eigenspace of   A   are   $\displaystyle \frac{n+1}{2}$   dimensional. Now, we choose   P   as in Case 3 with   $k=\displaystyle \frac{n+1}{2}$ variables.

Remark 4.10

Note that choosing an appropriate P depends on the number of $-1$ eigenvalues of the matrix A and it can be generalized to all dimensions.

The calculations above are done with the program Maple.