$ \newcommand\Cl{\mathrm{Cl} \newcommand\tr{\mathop{\mathrm{tr}}}} \newcommand\Ext{{\textstyle\bigwedge}} \newcommand\form[1]{\langle#1\rangle} \newcommand\Hom{\mathop{\mathrm{Hom}}} \newcommand\rev[1]{#1^\perp} \newcommand\doub\mathfrak $

Suppose $k$ has characteristic $\not= 2$ and that $V$ is finite dimensional (which I think you assumed implicitly). Let $V^*$ be the dual of $V$. We will also need to assume that $Q$ is nondegerate, i.e. its bilinear form gives an isomorphism $V \cong V^*$.

Then $\dim\Cl(V, Q) = 2^n$; if we define the trace of an element of $\Cl(V, Q)$ as the trace of its left multiplication $$ \tr X = \tr(Y \mapsto XY) $$ then your $\pi : \Cl(V, Q) \to k$ is exactly $$ \pi(X) = \frac1{2^n}\tr X. $$ This can be confirmed by using an orthonormal basis. Alternatively, we could construct the canonical isomorphism $\Cl(V, Q) \cong \Ext V$ where then $$ \pi(X) = \form{X}_0 $$ where the RHS is the projection onto the scalar (i.e. grade 0) part imported from $\Ext V$ into $\Cl(V,Q)$. Briefly, we can do this by defining $\wedge : V \times \Cl(V,Q) \to \Cl(V,Q)$ by $v\wedge X = \frac12(vX + \hat Xv)$ where $\hat X$ is the main involution applied to $X$ (i.e. the involution that negates all vectors); considering the map $V \to \Hom_k\Cl(V,Q)$ given by $v \mapsto v\wedge\cdot$, the universal property of $\Ext V$ extends this to an algebra homomorphism $\varphi : \Ext V \to \Hom_k\Cl(V, Q)$, and the map $X \mapsto \varphi(X)(1)$ can be shown to be the desired isomorphism. This is a particular instance of Chevalley's maps $\Cl(V, Q) \cong \Cl(V, Q+Q')$; see these notes by Darij Grinberg for a general approach descending from the tensor algebra.

The bilinear form associated to $Q$ $$ B(v, w) = \frac12(Q(v + w) - Q(v) - Q(w)) $$ induces an isomorphism $\flat : V \to V^*$; using this to apply $Q$ to $V^*$, the universal properties of the Clifford algebras $\Cl(V, Q), \Cl(V^*, Q)$ extend $\flat$ to an algebra isomorphism $\Cl(V, Q) \to \Cl(V^*, Q)$. Using the natural bilinear pairing on $\Ext V^*\times\Ext V \to k$ $$ (v^*_l\wedge v^*_{l-1}\wedge\cdots\wedge v^*_1,\; v_1\wedge v_2\wedge\cdots\wedge v_m) \mapsto \delta_{lm}\det\bigl(v^*_i(v_j)\bigr)_{i,j=1}^m $$ then induces a linear isomorphism $\Ext V^* \to (\Ext V)^*$, and composing with $\flat$ finally gives a linear isomorphism $\flat' : \Cl(V, Q) \to \Cl(V, Q)^*$; the bilinear form associated to this turns out to be exactly $$ (X, Y) \mapsto \form{XY}_0. $$ We can actually get the pairing $\Ext V^*\times\Ext V \to k$ by considering the natural Clifford algebra on $V^*\oplus V$ and using the trace definition of the scalar part, but I will not digress further.

Now we define the isomorphism $M^\vee \to M^*$ by $$ \psi \mapsto \Phi_\psi,\quad \Phi_\psi(m) = \form{\psi(m)}_0. $$ This is a homomorphism since $\form{XY}_0 = \form{YX}_0$ and so $$ \Phi_{\psi\cdot g}(m) = \form{\psi(m)g}_0 = \form{g\psi(m)}_0 = \form{\psi(gm)}_0 = (\Phi_\psi\cdot g)(m). $$ The inverse is given by $$ \phi \mapsto \Psi_\phi,\quad \Psi_\phi(m) = (g \mapsto \phi(gm))^{\sharp'} $$ where $\sharp' = (\flat')^{-1}$. Observe: $$ \Psi_{\Phi_\psi}(m) = (g \mapsto \Phi_\psi(gm))^{\sharp'} = (g \mapsto \form{\psi(gm)}_0)^{\sharp'} = (g \mapsto \form{g\psi(m)}_0)^{\sharp'} = \psi(m). $$

Your formula in terms of $G^+$ comes from expressing $\sharp'$ using that basis; the reciprocal basis of $G^+$ is $\{g^{-1} \;:\; g \in G^+\}$.

I have no idea how to handle characteristic 2. The main issue is that there is no canonical isomorphism with $\Ext V$ in this case, and we can't even use the trace since $\tr X = 0$ for all $X$.

$ \newcommand\Cl{\mathrm{Cl} \newcommand\tr{\mathop{\mathrm{tr}}}} \newcommand\Ext{{\textstyle\bigwedge}} \newcommand\form[1]{\langle#1\rangle} \newcommand\Hom{\mathop{\mathrm{Hom}}} \newcommand\rev[1]{#1^\perp} \newcommand\doub\mathfrak $Suppose $k$ has characteristic $\ne 2$ and that $V$ is finite dimensional (which I think you assumed implicitly). Let $V^*$ be the dual of $V$. We will also need to assume that $Q$ is nondegerate, i.e. its bilinear form gives an isomorphism $V \cong V^*$.

Then $\dim\Cl(V, Q) = 2^n$; if we define the trace of an element of $\Cl(V, Q)$ as the trace of its left multiplication $$ \tr X = \tr(Y \mapsto XY) $$ then your $\pi : \Cl(V, Q) \to k$ is exactly $$ \pi(X) = \frac1{2^n}\tr X. $$ This can be confirmed by using an orthonormal basis. Alternatively, we could construct the canonical isomorphism $\Cl(V, Q) \cong \Ext V$ where then $$ \pi(X) = \form{X}_0 $$ where the RHS is the projection onto the scalar (i.e. grade 0) part imported from $\Ext V$ into $\Cl(V,Q)$. Briefly, we can do this by defining $\wedge : V \times \Cl(V,Q) \to \Cl(V,Q)$ by $v\wedge X = \frac12(vX + \hat Xv)$ where $\hat X$ is the main involution applied to $X$ (i.e. the involution that negates all vectors); considering the map $V \to \Hom_k\Cl(V,Q)$ given by $v \mapsto v\wedge\cdot$, the universal property of $\Ext V$ extends this to an algebra homomorphism $\varphi : \Ext V \to \Hom_k\Cl(V, Q)$, and the map $X \mapsto \varphi(X)(1)$ can be shown to be the desired isomorphism. This is a particular instance of Chevalley's maps $\Cl(V, Q) \cong \Cl(V, Q+Q')$; see the notes The Clifford algebra and the Chevalley map — a computational approach by Darij Grinberg for a general approach descending from the tensor algebra.

The bilinear form associated to $Q$ $$ B(v, w) = \frac12(Q(v + w) - Q(v) - Q(w)) $$ induces an isomorphism $\flat : V \to V^*$; using this to apply $Q$ to $V^*$, the universal properties of the Clifford algebras $\Cl(V, Q), \Cl(V^*, Q)$ extend $\flat$ to an algebra isomorphism $\Cl(V, Q) \to \Cl(V^*, Q)$. Using the natural bilinear pairing on $\Ext V^*\times\Ext V \to k$ $$ (v^*_l\wedge v^*_{l-1}\wedge\dotsb\wedge v^*_1,\; v_1\wedge v_2\wedge\dotsb\wedge v_m) \mapsto \delta_{lm}\det\bigl(v^*_i(v_j)\bigr)_{i,j=1}^m $$ then induces a linear isomorphism $\Ext V^* \to (\Ext V)^*$, and composing with $\flat$ finally gives a linear isomorphism $\flat' : \Cl(V, Q) \to \Cl(V, Q)^*$; the bilinear form associated to this turns out to be exactly $$ (X, Y) \mapsto \form{XY}_0. $$ We can actually get the pairing $\Ext V^*\times\Ext V \to k$ by considering the natural Clifford algebra on $V^*\oplus V$ and using the trace definition of the scalar part, but I will not digress further.

Now we define the isomorphism $M^\vee \to M^*$ by $$ \psi \mapsto \Phi_\psi,\quad \Phi_\psi(m) = \form{\psi(m)}_0. $$ This is a homomorphism since $\form{XY}_0 = \form{YX}_0$ and so $$ \Phi_{\psi\cdot g}(m) = \form{\psi(m)g}_0 = \form{g\psi(m)}_0 = \form{\psi(gm)}_0 = (\Phi_\psi\cdot g)(m). $$ The inverse is given by $$ \phi \mapsto \Psi_\phi,\quad \Psi_\phi(m) = (g \mapsto \phi(gm))^{\sharp'} $$ where $\sharp' = (\flat')^{-1}$. Observe: $$ \Psi_{\Phi_\psi}(m) = (g \mapsto \Phi_\psi(gm))^{\sharp'} = (g \mapsto \form{\psi(gm)}_0)^{\sharp'} = (g \mapsto \form{g\psi(m)}_0)^{\sharp'} = \psi(m). $$

Your formula in terms of $G^+$ comes from expressing $\sharp'$ using that basis; the reciprocal basis of $G^+$ is $\{g^{-1} \;:\; g \in G^+\}$.

I have no idea how to handle characteristic 2. The main issue is that there is no canonical isomorphism with $\Ext V$ in this case, and we can't even use the trace since $\tr X = 0$ for all $X$.

$ \newcommand\Cl{\mathrm{Cl} \newcommand\tr{\mathop{\mathrm{tr}}}} \newcommand\Ext{{\textstyle\bigwedge}} \newcommand\form[1]{\langle#1\rangle} \newcommand\Hom{\mathop{\mathrm{Hom}}} \newcommand\rev[1]{#1^\perp} \newcommand\doub\mathfrak $

Suppose $k$ has characteristic $\not= 2$ and that $V$ is finite dimensional (which I think you assumed implicitly). Let $V^*$ be the dual of $V$.

Then $\dim\Cl(V, Q) = 2^n$; if we define the trace of an element of $\Cl(V, Q)$ as the trace of its left multiplication $$ \tr X = \tr(Y \mapsto XY) $$ then your $\pi : \Cl(V, Q) \to k$ is exactly $$ \pi(X) = \frac1{2^n}\tr X. $$ This can be confirmed by using an orthonormal basis. Alternatively, we could construct the canonical isomorphism $\Cl(V, Q) \cong \Ext V$ where then $$ \pi(X) = \form{X}_0 $$ where the RHS is the projection onto the scalar (i.e. grade 0) part imported from $\Ext V$ into $\Cl(V,Q)$. Briefly, we can do this by defining $\wedge : V \times \Cl(V,Q) \to \Cl(V,Q)$ by $v\wedge X = \frac12(vX + \hat Xv)$ where $\hat X$ is the main involution applied to $X$ (i.e. the involution that negates all vectors); considering the map $V \to \Hom_k\Cl(V,Q)$ given by $v \mapsto v\wedge\cdot$, the universal property of $\Ext V$ extends this to an algebra homomorphism $\varphi : \Ext V \to \Hom_k\Cl(V, Q)$, and the map $X \mapsto \varphi(X)(1)$ can be shown to be the desired isomorphism. This is a particular instance of Chevalley's maps $\Cl(V, Q) \cong \Cl(V, Q+Q')$; see these notes by Darij Grinberg for a general approach descending from the tensor algebra.

The bilinear form associated to $Q$ $$ B(v, w) = \frac12(Q(v + w) - Q(v) - Q(w)) $$ induces an isomorphism $\flat : V \to V^*$; using this to apply $Q$ to $V^*$, the universal properties of the Clifford algebras $\Cl(V, Q), \Cl(V^*, Q)$ extend $\flat$ to an algebra isomorphism $\Cl(V, Q) \to \Cl(V^*, Q)$. Using the natural bilinear pairing on $\Ext V^*\times\Ext V \to k$ $$ (v^*_l\wedge v^*_{l-1}\wedge\cdots\wedge v^*_1,\; v_1\wedge v_2\wedge\cdots\wedge v_m) \mapsto \delta_{lm}\det\bigl(v^*_i(v_j)\bigr)_{i,j=1}^m $$ then induces a linear isomorphism $\Ext V^* \to (\Ext V)^*$, and composing with $\flat$ finally gives a linear isomorphism $\beta' : \Cl(V, Q) \to \Cl(V, Q)^*$; the bilinear form associated to this turns out to be exactly $$ (X, Y) \mapsto \form{XY}_0. $$ We can actually get the pairing $\Ext V^*\times\Ext V \to k$ by considering the natural Clifford algebra on $V^*\oplus V$ and using the trace definition of the scalar part, but I will not digress further.

Now we define the isomorphism $M^\vee \to M^*$ by $$ \psi \mapsto \Phi_\psi,\quad \Phi_\psi(m) = \form{\psi(m)}_0. $$ This is a homomorphism since $\form{XY}_0 = \form{YX}_0$ and so $$ \Phi_{\psi\cdot g}(m) = \form{\psi(m)g}_0 = \form{g\psi(m)}_0 = \form{\psi(gm)}_0 = (\Phi_\psi\cdot g)(m). $$ The inverse is given by $$ \phi \mapsto \Psi_\phi,\quad \Psi_\phi(m) = (g \mapsto \phi(gm))^{\sharp'} $$ where $\sharp' = (\flat')^{-1}$. Observe: $$ \Psi_{\Phi_\psi}(m) = (g \mapsto \Phi_\psi(gm))^{\sharp'} = (g \mapsto \form{\psi(gm)}_0)^{\sharp'} = (g \mapsto \form{g\psi(m)}_0)^{\sharp'} = \psi(m). $$

Your formula in terms of $G^+$ comes from expressing $\sharp'$ using that basis; the reciprocal basis of $G^+$ is $\{g^{-1} \;:\; g \in G^+\}$.

I have no idea how to handle characteristic 2. The main issue is that there is no canonical isomorphism with $\Ext V$ in this case, and we can't even use the trace since $\tr X = 0$ for all $X$.

$ \newcommand\Cl{\mathrm{Cl} \newcommand\tr{\mathop{\mathrm{tr}}}} \newcommand\Ext{{\textstyle\bigwedge}} \newcommand\form[1]{\langle#1\rangle} \newcommand\Hom{\mathop{\mathrm{Hom}}} \newcommand\rev[1]{#1^\perp} \newcommand\doub\mathfrak $

Suppose $k$ has characteristic $\not= 2$ and that $V$ is finite dimensional (which I think you assumed implicitly). Let $V^*$ be the dual of $V$. We will also need to assume that $Q$ is nondegerate, i.e. its bilinear form gives an isomorphism $V \cong V^*$.

Then $\dim\Cl(V, Q) = 2^n$; if we define the trace of an element of $\Cl(V, Q)$ as the trace of its left multiplication $$ \tr X = \tr(Y \mapsto XY) $$ then your $\pi : \Cl(V, Q) \to k$ is exactly $$ \pi(X) = \frac1{2^n}\tr X. $$ This can be confirmed by using an orthonormal basis. Alternatively, we could construct the canonical isomorphism $\Cl(V, Q) \cong \Ext V$ where then $$ \pi(X) = \form{X}_0 $$ where the RHS is the projection onto the scalar (i.e. grade 0) part imported from $\Ext V$ into $\Cl(V,Q)$. Briefly, we can do this by defining $\wedge : V \times \Cl(V,Q) \to \Cl(V,Q)$ by $v\wedge X = \frac12(vX + \hat Xv)$ where $\hat X$ is the main involution applied to $X$ (i.e. the involution that negates all vectors); considering the map $V \to \Hom_k\Cl(V,Q)$ given by $v \mapsto v\wedge\cdot$, the universal property of $\Ext V$ extends this to an algebra homomorphism $\varphi : \Ext V \to \Hom_k\Cl(V, Q)$, and the map $X \mapsto \varphi(X)(1)$ can be shown to be the desired isomorphism. This is a particular instance of Chevalley's maps $\Cl(V, Q) \cong \Cl(V, Q+Q')$; see these notes by Darij Grinberg for a general approach descending from the tensor algebra.

The bilinear form associated to $Q$ $$ B(v, w) = \frac12(Q(v + w) - Q(v) - Q(w)) $$ induces an isomorphism $\flat : V \to V^*$; using this to apply $Q$ to $V^*$, the universal properties of the Clifford algebras $\Cl(V, Q), \Cl(V^*, Q)$ extend $\flat$ to an algebra isomorphism $\Cl(V, Q) \to \Cl(V^*, Q)$. Using the natural bilinear pairing on $\Ext V^*\times\Ext V \to k$ $$ (v^*_l\wedge v^*_{l-1}\wedge\cdots\wedge v^*_1,\; v_1\wedge v_2\wedge\cdots\wedge v_m) \mapsto \delta_{lm}\det\bigl(v^*_i(v_j)\bigr)_{i,j=1}^m $$ then induces a linear isomorphism $\Ext V^* \to (\Ext V)^*$, and composing with $\flat$ finally gives a linear isomorphism $\flat' : \Cl(V, Q) \to \Cl(V, Q)^*$; the bilinear form associated to this turns out to be exactly $$ (X, Y) \mapsto \form{XY}_0. $$ We can actually get the pairing $\Ext V^*\times\Ext V \to k$ by considering the natural Clifford algebra on $V^*\oplus V$ and using the trace definition of the scalar part, but I will not digress further.

Now we define the isomorphism $M^\vee \to M^*$ by $$ \psi \mapsto \Phi_\psi,\quad \Phi_\psi(m) = \form{\psi(m)}_0. $$ This is a homomorphism since $\form{XY}_0 = \form{YX}_0$ and so $$ \Phi_{\psi\cdot g}(m) = \form{\psi(m)g}_0 = \form{g\psi(m)}_0 = \form{\psi(gm)}_0 = (\Phi_\psi\cdot g)(m). $$ The inverse is given by $$ \phi \mapsto \Psi_\phi,\quad \Psi_\phi(m) = (g \mapsto \phi(gm))^{\sharp'} $$ where $\sharp' = (\flat')^{-1}$. Observe: $$ \Psi_{\Phi_\psi}(m) = (g \mapsto \Phi_\psi(gm))^{\sharp'} = (g \mapsto \form{\psi(gm)}_0)^{\sharp'} = (g \mapsto \form{g\psi(m)}_0)^{\sharp'} = \psi(m). $$

Your formula in terms of $G^+$ comes from expressing $\sharp'$ using that basis; the reciprocal basis of $G^+$ is $\{g^{-1} \;:\; g \in G^+\}$.

I have no idea how to handle characteristic 2. The main issue is that there is no canonical isomorphism with $\Ext V$ in this case, and we can't even use the trace since $\tr X = 0$ for all $X$.