Lemma 2

Let $K'$ be a non-discrete closed subfield of K, A a $K'$-manifold, B and C $K$-manifolds and $f$ a $K'$-analytic mapping of $A \times B$ into C. Suppose that, for all

LIE GROUPS

$a \in A$, the mapping $b \mapsto f(a, b)$ of B into C is K-*analytic*. Then, for all $t \in TA$, the mapping $u \mapsto (Tf)(t, u)$ of TB into TC is K-*analytic*.

We fix $t \in TA$ and write $g(u) = (Tf)(t, u)$. Clearly $g$ is K'-analytic. By Differentiable and Analytic Manifolds, R, 5.14.6, it suffices to prove that the tangent mappings to $g$ are K-linear. It can be assumed that A, B, C are open neighbourhoods of 0 in complete normable spaces E, F, G over K', K, K and that $t$ is tangent to A at 0. We identify TA, TB, TC, with $A \times E, B \times F, C \times G$ and $t$ with an element of E. Then for all $(x, y) \in TB = B \times F$.

We identify $T(B \times F)$ with $(B \times F) \times (F \times F)$ and $T(C \times G)$ with $(C \times G) \times (G \times G)$. Then, for all

$(Tg)((x, y), (h, k)) = ((a, b), (c, d))$, where

We now fix $(x, y) \in B \times F$. We need to prove that the mapping $(h, k) \mapsto (c, d)$ of $F \times F$ into $G \times G$ is K-linear. As the mapping $x \mapsto f(0, x)$ of B into C is K-analytic, the mappings

are K-linear. On the other hand

and, for fixed $\lambda$, the mapping $x \mapsto f(\lambda t, x)$ is K-analytic, so that the mapping $h \mapsto (D_2f)(\lambda t, x)(h)$ is K-linear.