Lemma 1

Let M be an analytic manifold over $k$ and $a= (a_0, . . . , a_{n-1}, a_n= 1)$ a sequence of analytic functions on M. For all $x\in M$, let $r_a(x)$ be the upper bound of those $i\in 0, n$ such that $a_j(x) = 0$ for $j < i$ and let $r_a^0(x)$ be the upper bound of those $i\in 0, n$ such that $a_j$ is zero on a neighbourhood of $x$ for $j < i$.

(i) The function $r_a$ is upper semi-continuous.

(ii) For all $x\in M,r_a^0(x) =$ lim inf$_{y\rightarrow x}r_a(y)$.

(iii) The function $r_a^0$ is locally constant.

(iv) The set of points $x\in M$ such that $r^0_a(x) =r_a(x)$ is the set of points of M on a neighbourhood of which $r_a$ is constant. This is a dense open subset of M. If $k=\mathbf{C}$ and M is finite dimensional and connected, it is open and connected.

(i) If $r_a(x) =i$, then $a_i(x)\not= 0$ and, for all $y$ in a neighbourhood of $x$, we have $a_i(y)\not= 0$, so $r_a(y)\leq i$.

(ii) If $r^0_a(x) =i$, the functions $a_0, . . . , a_{i-1}$ are zero on a neighbourhood of $x$ and, for any $y$ in this neighbourhood, $r_a(y)\geq i$. Consequently, lim inf$_{y\rightarrow x}r_a(y)\geq i$. Every neighbourhood of $x$ contains a point $y$ such that $a_i(y)\not= 0$ and hence $r_a(y)\leq i$. Thus lim inf$_{y\rightarrow x}r_a(y) =i$.

(iii) Let $i=r_a^0(x)$ and let V be a neighbourhood of $x$ such that $a_j(y) = 0$ for all $y\in V$ and all $j < i$. Then $x\in M$-- Z, where Z denotes the set of points of M in a neighbourhood of which the function $a_i$ is zero. Since Z is closed in M (Differentiable and Analytic Manifolds, Results$, 5.3.5$)$, V\cap (M$ -- Z) is a neighbourhood of $x$. For every point $y$ in this neighbourhood, $r_a^0(y) =i$.

(iv) The function $r_a-r_a^0$ is upper semi-continuous and its value at any point is $\geq 0$. If $r_a(x) =r^0_a(x),r_a-r^0_a$ is zero on a neighbourhood of $x$, which shows that $r_a$ is constant on a neighbourhood of $x$ by (iii). Conversely, if $r_a$ is constant on a neighbourhood of $x$, then $r_a^0(x) =r_a(x)$ by (ii). The set of points $x\in M$ such that $r_a^0(x) =r_a(x)$ is thus an open subset $\Omega$ of M. If $x\in M$ and if $r^0_a(x)< r_a(x)$, every neighbourhood of $x$ contains a point $y$ such that $r_a(y)< r_a(x)$ and $r^0_a(y) =r_a^0(x)$. Every neighbourhood of $x$ thus contains a point $y$ such that

It follows that $\Omega$ is dense in M.

If M is connected and if $p$ is the value of $r_a^0$ on M, the points of $\Omega$ are the points $x\in M$ such that $a_p(x)\not= 0$. If $k=\mathbf{C}$, this implies that $\Omega$ is connected by Lemma 3 of Appendix II.

Let $\rho$ be an analytic linear representation of G on a vector space V of finite dimension $n$ over $k$. Put

det(T $-\rho (g) + 1$) $=a_0(g) +a_1(g)T +\cdots +a_{n-1}(g)T^{n-1}+ T^n$.

The functions $r_a$ and $r_a^0$ associated to the sequence $(a_0, a_1, . . . , a_{n-1},1)$ will be denoted by $r_{\rho}$ and $r^0_{\rho}$, respectively. Then, for all $g\in G$,

$r_{\rho}(g) =$ dim $V^1(\rho (g))$

$r_{\rho}^0(g) =$ lim inf$_{g'\rightarrow g}$ dim $V^1(\rho (g'))$.