Lemma 3

Let $t$ be a tensor in $T_{n} \cap J$. The homogeneous component $t_{n}$ of $t$ of order $n$ is in the kernel $I$ of the canonical homomorphism $T \to S$.

We write $t_{n}$ in the form $\sum_{i=1}^{r} x_{M_{i}}$, where the $M_{i}$ are sequences of $n$ elements of $\Lambda$. The mapping $\sigma$ extends to a homomorphism of the algebra $T$ into the algebra $\mathscr{L}_{K}(P)$ (which we shall also denote by $\sigma$), which is zero on $J$. By Lemma 2, $\sigma(t) . 1$ is a polynomial whose terms of highest degree are $\sum_{i=1}^{r} z_{M_{i}}$. As $t \in J$, $\sigma(t) = 0$ and hence $\sum_{i=1}^{r} z_{M_{i}} = 0$ in $P$. Now $P$ is canonically identified with $S$, since $g$ has basis $(x_{\lambda})$. Hence the canonical image of $t_{n}$ in $S$ is zero, that is $t_{n} \in I$.

We can now prove Theorem 1. It is necessary to prove that the canonical homomorphism of $S$ onto $G$ is injective. In other words, if $t \in T^{n}$ and $\psi$ denotes the canonical homomorphism of $T$ onto $U$, it is necessary to show that the condition $\psi(t) \in U_{n-1}$ implies $t \in I$. Now $\psi(t) \in U_{n-1}$ means that there exists a tensor $t' \in T_{n-1}$ such that $t - t' \in J$. The tensor $t - t'$ admits $t$ as homogeneous component of order $n$ and hence $t \in I$ by Lemma 3.