Affine Weyl group is a coroot lattice semidirect product

Extend the $\nu$ in the statement to vanish on $c,d$. The displayed operators fix $\delta$ and preserve $\lambda (c)$. In a composition the additional cross term in the $\delta$ coefficient is $B(\gamma ,\eta )\lambda (c)$, so symmetry gives $t_{\gamma }{t}_{\eta }=t_{\gamma +\eta }$. Thus $t_0=1$ and $t_{-\gamma }$ is the inverse. If $t_{\gamma }=1$, applying it to every level-zero functional gives $\lambda (\gamma )=0$ for all finite $\lambda$, hence $\gamma =0$.

In the Euclidean coroot system let $V$ be the real span of $W_0{\theta }^{\vee }$. It is nonzero and $W_0$-invariant. For a coroot $\beta$ not orthogonal to $V$, some $v\in V$ has $(v,\beta )\mathrm{\ne }0$; the reflection formula makes $v-s_{\beta }v$ a nonzero multiple of $\beta$, so $\beta \in V$. Thus every coroot belongs to either $V$ or $V^{\perp }$. If both classes occurred, finite roots would likewise split into two nonempty orthogonal classes. The corresponding root spaces and coroot spans would form commuting ideals: a mixed root sum lies in neither span, so is not a root; mixed Cartan actions vanish; opposite-root brackets lie in their corresponding coroot spans. This contradicts simplicity of $\mathfrak{g}$. Consequently $V$ is the entire coroot space.

Finite $W_0$ fixes $c,d,\delta$, preserves $B$ and satisfies $w\nu (\gamma )=\nu (w\gamma )$. Substitution proves $w{t}_{\gamma }{w}^{-1}=t_{w\gamma }$. Since $\nu ({\theta }^{\vee })=\theta$ and $B({\theta }^{\vee },{\theta }^{\vee })=2$, substitution also gives $t_{{\theta }^{\vee }}{s}_{\theta }(\lambda )=\lambda -(\lambda (c)-\lambda ({\theta }^{\vee }))(\delta -\theta )=s_0\lambda \mathrm{.}$ Here $s_{\theta }\in W_0$ by F5. Hence $t_{{\theta }^{\vee }}$ and all its finite Weyl conjugates belong to $W_{\mathrm{a}\mathrm{f}\mathrm{f}}$.

Let $M$ be the integer span of $W_0{\theta }^{\vee }$. For any coroot $\beta$, step 1.2 supplies an orbit coroot $\alpha$ with $(\alpha ,\beta )\mathrm{\ne }0$. It has minimal coroot length because $\theta$ is long. If proportional, reducedness gives $\beta =\pm \alpha \in M$. Otherwise integrality and strict Cauchy–Schwarz give $0<\mid \frac{2(\alpha ,\beta )}{(\beta ,\beta )}\mid <\frac{2\mathrm{\parallel }\alpha \mathrm{\parallel }}{\mathrm{\parallel }\beta \mathrm{\parallel }}\le 2.$ The integer has absolute value one. Since $M$ is Weyl invariant, $\alpha -s_{\beta }\alpha =\pm \beta \in M$. Every coroot belongs to $M$, so $M=Q^{\vee }$. Steps 1.1 and 2.1 therefore put every $t_{\gamma }$, $\gamma \in Q^{\vee }$, in $W_{\mathrm{a}\mathrm{f}\mathrm{f}}$.

The products $t_{\gamma }w$ form a group by steps 1.1 and 2.1 and lattice preservation. They contain all finite simple reflections and $s_0=t_{{\theta }^{\vee }}{s}_{\theta }$, so form all of $W_{\mathrm{a}\mathrm{f}\mathrm{f}}$. Let $\mathrm{\Lambda }$ vanish on $\mathfrak{h},d$ and have $\mathrm{\Lambda }(c)=1$. Finite $W_0$ fixes it, whereas $t_{\gamma }\mathrm{\Lambda }=\mathrm{\Lambda }+\nu (\gamma )-B(\gamma ,\gamma )\delta \mathrm{/}2$. Thus $t_{\gamma }\in W_0$ forces $\nu (\gamma )=0$, hence $\gamma =0$. Trivial intersection gives uniqueness by comparing two products. Zero translation, identity finite factor and rank one are included. Every lattice expression is a finite sum and no AC is used.