Wisconsin Bourbaki Seminar

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 24 March 2014

Notes and References

This is an excerpt from notes of the Wisconsin Bourbaki Seminar, Fall 1993. The notes were written by Oliver Eng, Susan Hollingsworth, Mark Logan, Arun Ram and Louis Solomon.

§3

2. Let $V$ be a finite dimensional real vector space with an inner product. Let $F$ be a finite subgroup of the orthogonal group of $V$ generated by reflections. Let $\Lambda$ be a discrete subgroup of $V$ stable under $F\text{.}$ Let $W$ be the group of affine transformations of $V$ generated by $F$ and translations by vectors in $\Lambda \text{.}$ Let $ℌ$ be the set of hyperplanes $H$ in $V$ such that $s_H\in F$ and let $R$ be the set of $r\in \Lambda$ such that there exists a hyperplane $h\in ℌ$ such that $⟨r,h⟩=0\text{.}$
   1. Show that $W$ is generated by (affine) reflections if and only if $R$ generates $\Lambda$ as a $\mathbb{Z}\text{-module.}$
   2. Consider ${ℝ}^2$ with the scalar product $((x,y),(x\prime ,y\prime ))\mapsto xx\prime +yy\prime \text{.}$ Let $$e_1=(1,0),e_2=(-\frac{1}{2},\frac{\sqrt{3}}{2}),e_3=(-\frac{1}{2},-\frac{\sqrt{3}}{2}),{\Delta }_i=ℝe_i,$$ and let $F$ be the dihedral group generated by the $s_{{\Delta }_i}\text{.}$ Let $\Lambda$ be the discrete subgroup of ${ℝ}^2$ generated by the $e_i\text{.}$ The subgroup $\Lambda$ is stable under $F\text{.}$ Show that $W$ is not generated by reflections.
   
   Solution.
   1. $\Leftarrow \text{:}$ Assume that $R$ generates $\Lambda \text{.}$ $W$ is generated by $F$ and translations in $\Lambda \text{.}$ Translations in $\Lambda$ are generated by translations in $R\text{.}$ So it is sufficient to show that translations by vectors in $R$ are generated by reflections. Let $r\in R$ and let $t_r$ be the corresponding translation. Suppose that $r$ is orthogonal to $H\in ℌ\text{.}$ Then
      1. $t_r=\left(t_r{s}_H\right)s_H\text{.}$
      2. $t_r{s}_H$ is order $2,$ since $s_H{t}_r{s}_H=t_{s_{H}r}=t_{-r}$ and thus $t_r{s}_H{t}_r{s}_H=t_r{t}_{-r}=1\text{.}$
      3. $t_r{s}_H$ fixes the hyperplane $H+\frac{1}{2}r,$ since if $x$ is on the hyperplane $H+\frac{1}{2}r$ then $s_{H}x$ is a corresponding point on the hyperplane $H-\frac{1}{2}r$ and thus $t_r{s}_{H}x=x\text{.}$
      It follows from b) and c) that $t_r{s}_H$ is the reflection about the hyperplane $H+\frac{1}{2}r\text{.}$ So $t_r=s_{H+\frac{1}{2}r}{s}_H\text{.}$ Thus translations by elements of $R$ are generated by reflections.
      $\Rightarrow \text{:}$ Suppose that $W$ is generated by reflections. If $\lambda \in V,$ let $t_{\lambda }$ denote the translation by the vector $\lambda \text{.}$ Then if we let $T=\left\{t_{\lambda }\hspace{0.17em}\right|\hspace{0.17em}\lambda \in V\},$ $T$ is normalized by $O\left(V\right),$ and hence the group of affine transformations of $V$ is the semidirect product of $T$ by $O\left(V\right),$ since only the identity is both a translation and an element of $O\left(V\right)\text{.}$ Thus, since $\Lambda$ is stable under $F,$ every element of $W$ is uniquely expressible as $t_{\lambda }·g$ for some $g\in F$ and $\lambda \in \Lambda ,$ and hence $\left\{t_{\lambda }\hspace{0.17em}\right|\hspace{0.17em}\lambda \in \Lambda \}$ is normalized by $F\text{.}$ Now consider an affine reflection $\varphi :V\to V\text{.}$ The reflection $\varphi$ is determined by its fixed set, which is an affine hyperplane. This affine hyperplane may be written as $H_{\lambda }=\{\lambda +\mu \hspace{0.17em}|\hspace{0.17em}\mu \in H\},$ where $\lambda \in V,$ $H$ is a (linear) hyperplane, and $\lambda$ is perpendicular to $H\text{.}$ Then $V=ℝ\lambda \oplus H$ and $\varphi$ acts by $\varphi (a\lambda +\mu )=(2-a)\lambda +\mu$ for $a\in ℝ,$ $\mu \in H$ (To get from $a\lambda +\mu$ to $\lambda +\mu ,$ we add $(1-a)\lambda \text{;}$ then add it again to get its mirror image. This is since the orthogonal projection of $a\lambda +\mu$ in the affine hyperplane is simply $\lambda +\mu \text{.}$ See the diagram.) But if we let $s_H\in O\left(V\right)$ be the orthogonal reflection in $H,$ then for $a\in ℝ,$ $\mu \in H,$ $$\begin{array}{ccc}{t}_{2\lambda }{s}_H(a\lambda +\mu )& =& t_{2\lambda }(-a\lambda +\mu )\\ & =& (2-a)\lambda +\mu \\ & =& \varphi (a\lambda +\mu )\text{.}\end{array}$$ So it follows that any affine reflection in $W$ is uniquely expressible as $t_{\lambda }{s}_H$ where $\lambda \in \Lambda ,$ $H\in ℌ,$ and $\lambda$ is orthogonal to $H$ (and hence $\lambda \in R\text{).}$ Now suppose $\lambda \in \Lambda$ is arbitrary. Then $t_{\lambda }\in W,$ may be written as a product of affine reflections in $W,$ say $$\begin{array}{cc}{t}_{\lambda }=t_{{\mu }_1}{g}_1{t}_{{\mu }_2}{g}_2\cdots t_{{\mu }_r}{g}_r,& (*)\end{array}$$ with ${\mu }_i\in \Lambda ,$ $g_i\in F,$ and where $t_{{\mu }_i}{g}_i$ is an affine reflection. By the remark above it follows that ${\mu }_i\in R$ for all $i=1,\dots ,r\text{.}$ Now note that if $\mu \in V,$ $g\in O\left(V\right),$ then $$g{t}_{\mu }=t_{g\mu }g\text{.}$$ Applying this repeatedly to the right hand side of $(*)$ we get $$\begin{array}{ccc}{t}_{\lambda }& =& t_{{\mu }_1}{t}_{g_1{\mu }_2}{t}_{g_1{g}_2{\mu }_3}\cdots t_{g_1{g}_2\cdots g_{r-1}{\mu }_r}{g}_1\cdots g_r\\ & =& t\nu g_1\cdots g_r,\end{array}$$ where $\nu ={\mu }_1+g_1{\mu }_2+\cdots +\cdots g_1\cdots g_{r-1}{\mu }_r\text{.}$ Thus $\lambda =\nu ,$ and $g_1\cdots g_r=1$ (which is not used). But $\Lambda$ is clearly stable under $W,$ and $W$ permutes $ℌ,$ and $W$ is orthogonal. Thus it folllows from the definition of $R$ that $WR\subseteq R\text{.}$ Therefore, since ${\mu }_i\in R$ for $i=1,\dots ,r,$ each of ${\mu }_1,g_1{\mu }_2,\dots ,g_1\cdots g_{r-1}{\mu }_r\in R\text{.}$ Thus $\lambda =\nu$ lies in the $\mathbb{Z}\text{-span}$ of elements of $R,$ and hence $R$ generates $\lambda \text{.}$
   2. See attached picture. $◻$
   
   
3. Let $V$ be a finite dimensional vector space over $ℝ,$ $W$ a finite subgroup of $GL\left(V\right)$ generated by reflections. Show that every element of order $2$ in $W$ is generated by pairwise commuting reflections belonging to $W\text{.}$
   
   Solution. The proof is by induction on the dimension of $V\text{.}$ If $\text{dim}\hspace{0.17em}V=1$ then there is only one element of order $2$ in $W$ and this is certainly a reflection. Assume $\text{dim}\hspace{0.17em}V>1$ and let $w\in W$ be an element of order $2\text{.}$ We may choose a basis of $V$ such that the matrix of $V$ with respect to this basis is diagonal with digonal entries $(-1,-1,\dots ,-1,1,1,\dots ,1)\text{.}$ There are two cases:
   Case 1. $w$ fixes a nonzero subspace $E$ of $V$ pointwise.
   Let $ℌ$ be the set of hyperplanes such that $W$ is generated by the reflections in these hyperplanes. Define $W\left(E\right)=\{w\in W\hspace{0.17em}|\hspace{0.17em}w\hspace{0.17em}\text{fixes}\hspace{0.17em}E\hspace{0.17em}\text{pointwise}\}\text{.}$ Then, by Proposition 2, Chapt. V §3, $W\left(E\right)$ is generated by reflections in the hyperplanes in the set ${ℌ}_E=\{H\in ℌ\hspace{0.17em}|\hspace{0.17em}H\supseteq E\}\text{.}$ Since $w$ fixes $E,$ $w$ is an element of $W\left(E\right)$ and $W\left(E\right)$ acts on $V/E$ and $\text{dim}\hspace{0.17em}V/E<\text{dim}\hspace{0.17em}V\text{.}$ By induction we have that $w|E^{\perp }=t_1{t}_2\cdots t_r$ where $t_i$ are pairwise commuting reflections. If we let $s_i$ be the reflection which acts on $E$ by the identity and on $E^{\perp }$ as $t_i$ then $w=s_1{s}_2\cdots s_r\text{.}$ It is clear that since the reflections $t_1,t_2,\dots ,t_r$ are pairwise commuting so are the reflections $s_1,s_2,\dots ,s_r\text{.}$ The reflections $s_i,$ $1\le i\le r$ are elements of $W$ since they are reflections in hyperplanes in ${ℌ}_E\text{.}$
   Case 2. The only element which $w$ fixes is $0\in V\text{.}$
   Then $wv=-v$ for all $v\in V\text{.}$ Since $-1$ is in the center of $GL\left(V\right),$ $w$ is in the center of $W\text{.}$ Since $W$ is generated by reflections we can write $w=s_1\cdots s_r$ as a product of reflections. Since $w$ is in the center $$w=s_{r}w{s}_r^{-1}=s_r{s}_1\cdots s_{r-1}\text{.}$$ Let $w\prime =s_1\cdots s_{r-1}\text{.}$ Then $${\left(w\prime \right)}^2=s_1\cdots s_{r-1}{s}_r^2{s}_1\cdots s_{r-1}=w^2=1\text{.}$$ Since $w\prime \ne w,$ then $w\prime \ne -1$ and so by case 1 we may write $w\prime$ as a product of pairwise commuting reflections $w=t_1\cdots t_m\text{.}$ Since $$\begin{array}{c}w=t_1\cdots t_m{s}_r=t_i{t}_1\cdots {\hat{t}}_i\cdots t_m{s}_r=s_r{t}_i{t}_1\cdots {\hat{t}}_i\cdots t_m,\text{and}\\ w=s_r{t}_1\cdots t_m=s_r{t}_i{t}_1\cdots {\hat{t}}_i\cdots t_m,\end{array}$$ it follows that $s_r{t}_i=t_i{s}_r$ for each $i\text{.}$ So $w$ can be written as a product of pairwise commuting reflections. $◻$
   
   Notes. Recall that The longest element of the Weyl group $w_0$ is an element of order $2\text{.}$ It is interesting to write these elements as a product of pairwise commuting reflections.
   
   
4. Let $V$ be a finite dimensional real vector space and let $W$ be a finite subgroup of $GL\left(V\right)$ generated by reflections. Let $w\in W\text{.}$ Suppose that $V\prime$ is a subspace of $V$ stable under $W$ and let $k$ be the order of the restriction $w|V\prime$ of $w$ to $V\prime \text{.}$ Show that there exists $x\in W$ of order $k,$ leaving $V\prime$ stable and such that $x|V\prime =w|V\prime \text{.}$
   
   Solution. Let $W\prime =\{w\in W\hspace{0.17em}|\hspace{0.17em}wv=v,\hspace{0.17em}\text{for all}\hspace{0.17em}v\in V\prime \}\text{.}$ By Proposition 2 §3.3, $W\prime$ is generated by reflections. Let $C$ be a chamber of $W\prime \text{.}$ Let $C\prime =wC,$ $w\in W,$ be another chamber. By Lemma 2 §3.1, there is an $h\in W\prime$ such that $hC\prime =C\text{.}$ So $whC\prime =C\prime \text{.}$ Then
   1. $wh$ stabilizes $V\prime$ since both $w$ and $h$ stabilize $V\prime \text{.}$
   2. $wh|V\prime =w|V\prime$ since $h$ fixes $V\prime$ pointwise.
   3. It remains to show that $wh$ has order $k\text{.}$
   Suppose that ${\left(wh\right)}^{\ell }=1\text{.}$ Then, since $wh|V\prime =w|V\prime$ we have that $${\left(wh\right)}^{\ell }|V\prime =w^{\ell }|V\prime =1|V\prime \text{.}$$ Since the order of $w|V\prime$ is $k$ it follows that $\ell \ge k\text{.}$ We will show that ${\left(wh\right)}^k=1\text{.}$ Since $wh$ stabilizes a chamber of $W\prime$ so does ${\left(wh\right)}^k\text{.}$ Since $wh|V\prime =w|V\prime$ and $w^k$ is the identity on $V\prime$ it follows that ${\left(wh\right)}^k$ is the identity on $V\prime \text{.}$ So ${\left(wh\right)}^k\in W\prime \text{.}$ Since ${\left(wh\right)}^k\in W\prime$ and ${\left(wh\right)}^k$ stabilizes a chamber of $W\prime$ it follows from Theorem 1 §3.2, that ${\left(wh\right)}^k$ is the identity in $W\prime \text{.}$ Therefore ${\left(wh\right)}^k$ is the identity in $W\text{.}$ $◻$
   
6. 1. Let $K$ be a commutative field and let $V$ be an $n$ dimensional vector space over $K\text{.}$ Let $\varphi$ be a symmetric bilinear form on $V$ and let $N=\{n\in V\hspace{0.17em}|\hspace{0.17em}\varphi (n,v)=0,\hspace{0.17em}\text{for all}\hspace{0.17em}v\in V\}$ be the null space of $\varphi \text{.}$ Suppose that $\text{dim}\hspace{0.17em}N=1\text{.}$ Show that the null space of the extension of $\varphi$ to ${\bigwedge }^{n-1}V$ is dimension $n-1\text{.}$
   2. Suppose that $K=ℝ$ and that $\varphi$ is positive. Let $(e_1,\dots ,e_n)$ be a base of $V,$ and let $a_{ij}=\varphi (e_i,e_j)\text{.}$ Suppose that $a_{ij}\le 0$ for $i\ne j\text{.}$ Suppose that $\{1,2,\dots ,n\}$ does not admit a partition $I\cup J$ such that $a_{ij}=0$ for all $i\in I$ and $j\in J\text{.}$ Let $A_{ij}$ be the cofactor of $a_{ij}$ in the matrix $\left(a_{ij}\right)\text{.}$ Show that $A_{ij}>0$ for all $i,j\text{.}$
   3. Let ${\eta }_1{e}_1+\cdots +{\eta }_n{e}_n$ be a vector with all coordinates $>0$ which generates the null space $N\text{.}$ Show that ${\eta }_1,\dots ,{\eta }_n$ are proportional to $\sqrt{A_{11}},\dots ,\sqrt{A_{nn}}\text{.}$
   
   Solution.
   1. The extension of the bilinear form $\varphi$ on $V$ to ${\bigwedge }^{k}V$ is given by, Alg. Chapt III §11.5 formula 30, $$\varphi (v_{i_1}\wedge \cdots \wedge v_{i_k},v_{i_1}^{\prime }\wedge \cdots \wedge v_{i_k}^{\prime })=\text{det}\left(\varphi (v_{i_j},v_{i_k})\right)\text{.}$$ Let $v_1\in N$ and complete this to a basis $v_1,\dots ,v_n$ of $V\text{.}$ The set of vectors $\{v_1\wedge \cdots \wedge {\hat{v}}_i\wedge \cdots \wedge v_n\}$ is a basis of ${\bigwedge }^{n-1}V\text{.}$ Let $M$ be the $n\times n$ matrix given by $M=\left(\varphi (v_k,v_{\ell })\right)$ and let $M_{ij}$ denote the matrix $M$ with the $i\text{th}$ row and the $j\text{th}$ column removed. Then for any $i,j$ such that $i\ne 1,$ $$\varphi (v_1\wedge \cdots \wedge {\hat{v}}_i\wedge \cdots \wedge v_n,v_1\wedge \cdots \wedge {\hat{v}}_j\wedge \cdots \wedge v_n)=\text{det}\left(M_{ij}\right)=0,$$ since $M_{ij}$ is a matrix with top row containing all zeros. It follows that all the basis vectors $v_1\wedge \cdots \wedge {\hat{v}}_i\wedge \cdots \wedge v_n$ are in the null space of the form on ${\bigwedge }^{n-1}V\text{.}$ Furthermore, since $\text{rank}\hspace{0.17em}M=\text{dim}\hspace{0.17em}N=1$ and $M$ is a matrix such that the first row and the first column are zero, $$\varphi (v_2\wedge \cdots \wedge v_n,v_2\wedge \cdots \wedge v_n)=\text{det}\left(M_{11}\right)\ne 0\text{.}$$ Thus the vector $v_2\wedge \cdots \wedge v_n$ is not an element of the null space of the form on ${\bigwedge }^{n-1}V\text{.}$
   2. Let $M=\left(\varphi (e_k,e_j)\right)=\left(a_{k_j}\right)$ and $M_{ij}$ be the matrices given in the proof of part a). Then $A_{ij}={(-1)}^{i+j}\text{det}\left(M_{ij}\right)\text{.}$ Then for any basis vector $e_k,$ we have that $$\begin{array}{ccc}\varphi (e_k,\sum _j{A}_{ij}{e}_j)& =& \sum _j{A}_{ij}\varphi (e_k,e_j)\\ & =& \sum _j{A}_{ij}{a}_{jk}\\ & =& {\delta }_{ik}\text{det}\left(M\right),\end{array}$$ by Cramer’s rule. Since $\text{det}\left(M\right)=0$ we have that $\varphi (e_k,\sum _j{A}_{ij}{e}_j)$ for all $k\text{.}$ It follows that $\sum _j{A}_{ij}{e}_j$ is an element of $N\text{.}$ By Lemma 4 §3.5 it follows that $N$ is spanned by a vector ${\eta }_1{e}_1+\cdots +{\eta }_n{e}_n$ such that ${\eta }_i>0$ for all $i\text{.}$ Since $\sum _j{A}_{ij}{e}_j\in N$ it follows that $A_{ij}=r_i{\eta }_j$ for some constants $r_i\text{.}$ A similar argument shows that $A_{ij}=c_j{\eta }_i$ for some constants $c_j\text{.}$ Since $A_{ii}=r_i{\eta }_i=c_i{\eta }_i$ and the ${\eta }_i>0,$ $r_i=c_i$ for all $i\text{.}$ Then since $A_{ij}=r_i{\eta }_j=r_j{\eta }_i,$ $$\frac{r_i}{{\eta }_i}=\frac{r_j}{{\eta }_j},\text{for all}\hspace{0.17em}i,j\text{.}$$ Setting $\mu =\frac{r_1}{{\eta }_1}$ we have that $$A_{ij}=\frac{r_i}{{\eta }_i}{\eta }_i{\eta }_j=\mu {\eta }_i{\eta }_j,$$ for all $i$ and $j\text{.}$ Since $\varphi$ is positive semidefinite we know that $A_{11}=\text{det}\left(M_{11}\right)\ge 0\text{.}$ We have already seen in part a) that $\text{det}\left(M_{11}\right)\ne 0\text{.}$ So $A_{11}=\mu {\eta }_1^2>0$ and we have that $\mu >0\text{.}$ It follows that $A_{ij}=\mu {\eta }_i{\eta }_j>0$ for all $i,j\text{.}$
   3. From the proof of part b) we have that $A_{ii}=\mu {\eta }_i^2$ where $\mu >0\text{.}$ It follows that $${\eta }_i=\frac{1}{\sqrt{\mu }}\sqrt{A_{ii}}\text{.}$$ So ${\eta }_i$ is proportional to $\sqrt{A_{ii}}\text{.}$ $◻$
   
   
7. Let $q({\xi }_1,\dots ,{\xi }_n)=\sum _{i,j}{a}_{ij}{\xi }_i{\xi }_j(a_{ij}=a_{ji})$ be a positive degenerate quadratic form on ${ℝ}^n,$ such that $a_{ij}\le 0$ for $i\ne j\text{.}$ Suppose that $\{1,2,\dots ,n\}$ does not admit a partition $I\cup J$ such that $a_{ij}=0$ for $i\in I,$ $j\in J\text{.}$
   1. Show that, if one puts $\xi =0,$ one gets a positive nondegenerate form by restricting to the coordinates ${\xi }_1,\dots ,{\xi }_{i-1},{\xi }_{i+1},\dots ,{\xi }_n\text{.}$
   2. Show that $a_{ii}>0$ for all $i\text{.}$
   3. Show that if one replaces one of the $a_{ij}$ with a value $a_{ij}^{\prime }\le a_{ij},$ the new form is nonpositive.
   
   Solution.
   1. By Lemma 4 §3.5, we have that the subspace of isotropic vectors is dimension $1$ and that this space is generated by a vector $\eta =({\eta }_1,\dots ,{\eta }_n)$ such that ${\eta }_i>0$ for all $i\text{.}$ If the form on ${ℝ}^{n-1}$ given by $q({\xi }_1,\dots ,{\xi }_{i-1},0,{\xi }_{i+1},\dots ,{\xi }_n)$ is degenerate then there is a nonzero vector $\xi =\left({\xi }_1,\dots ,{\xi }_{i-1},0,{\xi }_{i+1},\dots ,{\xi }_n\right)$ such that $q\left(\xi \right)=0\text{.}$ The vector $\xi$ must be a multiple of $\eta \text{.}$ This is a contradiction to the fact that $\xi$ is nonzero.
   2. Since the $q$ is positive we have that, for each $i,$ $0\le q\left({\epsilon }_i\right)=a_{ii},$ where ${\epsilon }_i$ is the vector with $i^{\text{th}}$ coordinate $1$ and all other coordinates $0\text{.}$ If $a_{ii}=0=q\left({\epsilon }_i\right),$ then by a) we must have that ${\epsilon }_i=0\text{.}$ This is a contradiction. So $a_{ii}>0$ for all $i\text{.}$
   3. Let $\eta =({\eta }_1,\dots ,{\eta }_n)$ be a vector such that ${\eta }_i>0$ for all $i$ and such that $q\left(\eta \right)=0\text{.}$ Then, if the new form is positive we have that $$0=\sum _{i,j}{a}_{ij}{\eta }_i{\eta }_j>\sum _{i,j}{a}_{ij}^{\prime }{\eta }_i{\eta }_j\ge 0,$$ which is clearly a contradiction. $◻$
   
   Notes. The null space of a form $\varphi$ is defined to be $$N=\{n\in V\hspace{0.17em}|\hspace{0.17em}\varphi (n,v)=0,\hspace{0.17em}\text{for all}\hspace{0.17em}v\in V\}\text{.}$$ Let $I$ be the set of isotropic vectors for $\varphi ,$ $$I=\{v\in V\hspace{0.17em}|\hspace{0.17em}\varphi (v,v)=0\}\text{.}$$ It is clear that if $n\in N$ the $n\in I$ since $\varphi (n,n)=0\text{.}$ So $N\subseteq I\text{.}$ Suppose that the form $\varphi$ is positive and let $n\in I\text{.}$ Let $v\in V$ and let $\lambda \in ℝ\text{.}$ Then $$\begin{array}{ccc}0& \le & \varphi (\lambda n+a,\lambda n+a)\\ & =& {\lambda }^2\varphi (n,n)+2\lambda \varphi (n,v)+\varphi (v,v)\\ & =& 0+2\lambda \varphi (n,v)+\varphi (v,v)\text{.}\end{array}$$ If $\varphi (n,v)\ne 0$ then we may choose $\lambda$ such that $2\lambda \varphi (v,n)<\varphi (v,v)\text{.}$ So $\varphi (n,v)=0\text{.}$ It follows that $n\in N\text{.}$ So $I=N$ if $\varphi$ is positive.
   
   
8. Let $\left(a_{ij}\right)$ be a real symmetric matrix with $n$ rows and $n$ columns.
   1. Put $s_k=\sum _{i=1}^k{a}_{ik}\text{.}$ For all ${\xi }_1,\dots ,{\xi }_n\in ℝ,$ one has that $$\sum _{i,k}{a}_{ik}{\xi }_i{\xi }_k=\sum _k{s}_k{\xi }_k^2-\frac{1}{2}\sum _{i,k}{a}_{ik}{({\xi }_i-{\xi }_k)}^2\text{.}$$
   2. Let ${\eta }_1,\dots ,{\eta }_n\in {ℝ}^{*}\text{.}$ Set $\sum _i{\eta }_i{a}_{ik}=t_k\text{.}$ Then $$\sum _{i,k}{a}_{ik}{\xi }_i{\xi }_k=\sum _k\frac{t_k{\xi }_k^2}{{\eta }_k}-\frac{1}{2}\sum _{i,k}{\eta }_i{\eta }_k{a}_{ik}{(\frac{{\xi }_i}{{\eta }_i}-\frac{{\xi }_k}{{\eta }_k})}^2\text{.}$$
   3. If there exist numbers ${\eta }_1,\dots ,{\eta }_n>0$ such that $\sum _i{\eta }_i{a}_{ik}=0$ $(k=1,2,\dots ,n),$ and if $a_{ij}\le 0$ for $i\ne j,$ then the quadratic form $\sum _{i,k}{a}_{ik}{\xi }_i{\xi }_k$ is positive degenerate.
   4. Let $\sum _{i,j}{q}_{ij}{\xi }_i{\xi }_k$ be a quadratic form on ${ℝ}^n$ such that $q_{ij}\le 0$ for $i\ne j\text{.}$ Suppose that $\{1,2,\dots ,n\}$ does not admit a partition $I\cup J$ such that $q_{ij}=0$ for $i\in I$ and $j\in J\text{.}$ Show that this form is positive degenerate if and only if there exist ${\eta }_1>0,\dots ,{\eta }_n>0$ such that $\sum _i{\eta }_i{q}_{ik}=0$ $(k=1,\dots ,n)\text{.}$
   
   Solution.
   1. Begin with the right hand side. $$\begin{array}{ccc}\sum _k{s}_k{\xi }_k^2-\frac{1}{2}\sum _{i,k}{a}_{ik}{({\xi }_i-{\xi }_k)}^2& =& \sum _{i,k}{a}_{ik}{\xi }_k^2-\frac{1}{2}\left(\sum _{i,k}{a}_{ik}({\xi }_i^2-2{\xi }_i{\xi }_j+{\xi }_j^2)\right)\\ & =& \sum _{i,k}{a}_{ik}{\xi }_k^2-\frac{1}{2}\sum _{i,k}{a}_{ik}{\xi }_i^2+\sum _{i,k}{a}_{ik}{\xi }_i{\xi }_k-\frac{1}{2}\sum _{i,k}{a}_{ik}{\xi }_k^2\\ & =& \frac{1}{2}\sum _{i,k}{a}_{ik}{\xi }_k^2-\frac{1}{2}\sum _{i,k}{a}_{ik}{\xi }_i^2+\sum _{i,k}{a}_{ik}{\xi }_i{\xi }_k\\ & =& \frac{1}{2}\sum _{i,k}(a_{ik}-a_{ki}){\xi }_k^2+\sum _{i,k}{a}_{ik}{\xi }_i{\xi }_k\\ & =& \sum _{i,k}{a}_{ik}{\xi }_i{\xi }_k,\end{array}$$ since $\left(a_{ij}\right)$ is symmetric.
   2. Replace ${\xi }_i$ by ${\xi }_i/{\eta }_i$ and $a_{ik}$ by ${\eta }_i{\eta }_k{a}_{ik}$ in a). Then $$s_k=\sum _i{\eta }_i{\eta }_k{a}_{ik}={\eta }_k\sum _i{\eta }_i{a}_{ik}={\eta }_k{t}_k\text{.}$$ Substituting, the formula in a) is $$\sum _{i,k}{\eta }_i{\eta }_k{a}_{ik}\frac{{\xi }_i}{{\eta }_i}\frac{{\xi }_k}{{\eta }_k}=\sum _k{\eta }_k{t}_k\frac{{\xi }_k^2}{{\eta }_k^2}-\frac{1}{2}\sum _{i,k}{\eta }_i{\eta }_k{a}_{ik}{(\frac{{\xi }_i}{{\eta }_i}-\frac{{\xi }_k}{{\eta }_k})}^2\text{.}$$ The trivial cancellations give the desired identity.
   3. Since $t_k=\sum _i{a}_{ik}{\eta }_k=0$ the identity in b) reduces to $$\begin{array}{ccc}\sum _{i,k}{a}_{ik}{\xi }_i{\xi }_k& =& -\frac{1}{2}\sum _{i,k}{\eta }_i{\eta }_k{a}_{ik}{(\frac{{\xi }_i}{{\eta }_i}-\frac{{\xi }_k}{{\eta }_k})}^2\\ & =& -\frac{1}{2}\sum _{i\ne k}{\eta }_i{\eta }_k{a}_{ik}{(\frac{{\xi }_i}{{\eta }_i}-\frac{{\xi }_k}{{\eta }_k})}^2\text{.}\end{array}$$ Since $a_{ik}\le 0$ for $i\ne k$ and ${\eta }_i>0$ for all $i$ we have that the right hand side is $\ge 0\text{.}$ It follows that the form is positive. The form is degenerate since $\sum _{i,k}{\eta }_k{\eta }_i{a}_{ik}=\sum _k{\eta }_k\sum _i{\eta }_i{a}_{ik}=0\text{.}$
   4. $\Leftarrow$ follows immediately from c).
      $\Rightarrow \text{:}$ In view of Lemma 4 §3.5 it is sufficient to show that if $\eta =({\eta }_1,\dots ,{\eta }_n)$ is a vector such that $q\left(\eta \right)=0$ and that ${\eta }_i>0$ for all $i$ then $\sum _i{\eta }_i{q}_{ik}=0$ $(k=1,\dots ,n)\text{.}$ Let $Q=\left(q_{ij}\right)\text{.}$ Let $\lambda \in ℝ$ and $x=(x_1,\dots ,x_n)\in {ℝ}^n\text{.}$ Then, since the form $q$ is positive, $$0\le \sum _{i,k}(x_i+\lambda {\eta }_i)(x_k+\lambda {\eta }_k)q_{ik}=\sum _{i,k}{x}_i{x}_k{q}_{ik}+2\lambda \sum _{i,k}{x}_k{\eta }_i{q}_{ik}+0$$ for arbitrary values of the $x_i$ and $\lambda \text{.}$ This is only possible if $\sum _{i,k}{x}_k{x}_i{q}_{ik}=0$ for all values of $x\text{.}$ So $$\sum _i{\eta }_i{q}_{ik}=0\text{.}◻$$
   
   Notes. Parts 8a) and 8b) are known as Crosby’s Lemma and are proved on pp. 177-178 of the book Regular Polytopes by H.S.M. Coxeter. See also the historical remarks pp. 185.

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