$\DeclareMathOperator\Cl{Cl}$Let $\Cl(V)$ be the Clifford algebra of a quadratic vector space $(V,Q)$ over $k$, and let $M$ be a (left) $\Cl(V)$-module. The dual Clifford module is typically defined as the linear dual $M^\ast=\operatorname{Hom}_k(M,k)$ with the Clifford action given by defining $g\cdot \phi$ as $g\cdot \phi\colon m\mapsto \phi(g^{\perp}m)$, where $\perp$ is the standard antiinvolution of $\Cl(V)$ given by $(v_1v_2\dotsm v_n)^\perp=v_n\dotsm v_2v_1$. This is similar (not to say it is the same) to the dual represetation of a finite group $G$, where the standard antiinvolution of $kG$ is $g\mapsto g^{-1}$.

Yet there is another, a priori distinct and actually more natural, definition of a dual Clifford module: one can consider ${M^\vee}=\operatorname{Hom}_{\Cl(V)}(M,\Cl(V))$ with its natural right $\Cl(V)$-module structure.

This is equivalently a left $\Cl(V)^\text{opp}$-module structure, and since $\perp$ is an algebra isomorphism from $\Cl(V)$ to $\Cl(V)^\text{opp}$, we see that $M^\vee=\operatorname{Hom}_{\Cl(V)}(M,\Cl(V))$ carries a left $\Cl(V)$-module structure via $\perp$.

Now, one may wonder whether $M^\ast$ and $M^\vee$ are isomorphic as left $\Cl(V)$-modules. Equivalently, and more naturally, one is wondering whether $M^\ast$ and $M^\vee$ are isomorphic as right $\Cl(V)$-modules, where the right $\Cl(V)$-module structure on $M^\ast$ is given by $\phi\cdot g\colon m\mapsto \phi(gm)$.

For the representations of a finite group $G$ this is precisely what happens: $M^\ast$ and $M^\vee=\operatorname{Hom}_{kG}(M,kG)$ are isomorphic as right $kG$-modules. An explicit isomorphism $M^\ast\to M^\vee$ is given by

$$\phi\mapsto\left(m \mapsto \sum_{g\in G} \phi(g^{-1}m)g\right).$$

The inverse isomorphism is simply picking the coefficient of the unit element of $G$.

For Clifford algebras things should go the same way, at least over $\mathbb{R}$ and $\mathbb{C}$. Let $(e_1,\dotsc, e_n)$ be an orthonormal basis of $V$, and let $G$ be the subgroup of $Pin(V)$ consisting of the elements $\{\pm e_1^{\epsilon_1}\dotsm e_n^{\epsilon_n}\}$ where $\epsilon_i\in \{0,1\}$. Then $\Cl(V)$-modules are equivalently $G$-modules such that the element $-1$ of $G$ acts as the multiplication by $-1$. One can then define an isomorphism $M^\ast\to M^\vee$ by

$$\phi\mapsto\left(m \mapsto \frac{1}{2}\sum_{g\in G} \phi(g^{-1}m)g\right)=\left(m \mapsto \sum_{g\in G^+} \phi(g^{-1}m)g\right),$$

where $G^+\subset G$ is the subset $\{e_1^{\epsilon_1}\dotsm e_n^{\epsilon_n}\}$ of “positive” elements. Here the sums are done in $\Cl(V)$. To write the inverse isomorphism one notices that there exist a linear morphism $\pi\colon \Cl(V)\to k$ that sends $1$ to $1$ and all other elements of $G^+$ to $0$.

Now the questions (assuming all of the above is correct), are:

i) can one give the isomorphism of left Clifford modules $M^\ast\cong M^\vee$ in an intrinsic way, without using an orthonormal basis (this would possibly give an isomorphism over an arbitrary field $k$)?

ii) if the intrinsic approach fails, can one at least show that the isomorphism above is independent of the chosen orthonormal basis? (This should be a simple check, but I'm postponing it in the hope of a positive answer to i).)

$\DeclareMathOperator\Cl{Cl}$Let $\Cl(V)$ be the Clifford algebra of a quadratic vector space $(V,Q)$ over $k$, and let $M$ be a (left) $\Cl(V)$-module. The dual Clifford module is typically defined as the linear dual $M^\ast=\operatorname{Hom}_k(M,k)$ with the Clifford action given by defining $g\cdot \phi$ as $g\cdot \phi\colon m\mapsto \phi(g^{\perp}m)$, where $\perp$ is the standard antiinvolution of $\Cl(V)$ given by $(v_1v_2\dotsm v_n)^\perp=v_n\dotsm v_2v_1$. This is similar (not to say it is the same) to the dual representation of a finite group $G$, where the standard antiinvolution of $kG$ is $g\mapsto g^{-1}$.

Yet there is another, a priori distinct and actually more natural, definition of a dual Clifford module: one can consider ${M^\vee}=\operatorname{Hom}_{\Cl(V)}(M,\Cl(V))$ with its natural right $\Cl(V)$-module structure.

This is equivalently a left $\Cl(V)^\text{opp}$-module structure, and since $\perp$ is an algebra isomorphism from $\Cl(V)$ to $\Cl(V)^\text{opp}$, we see that $M^\vee=\operatorname{Hom}_{\Cl(V)}(M,\Cl(V))$ carries a left $\Cl(V)$-module structure via $\perp$.

Now, one may wonder whether $M^\ast$ and $M^\vee$ are isomorphic as left $\Cl(V)$-modules. Equivalently, and more naturally, one is wondering whether $M^\ast$ and $M^\vee$ are isomorphic as right $\Cl(V)$-modules, where the right $\Cl(V)$-module structure on $M^\ast$ is given by $\phi\cdot g\colon m\mapsto \phi(gm)$.

For the representations of a finite group $G$ this is precisely what happens: $M^\ast$ and $M^\vee=\operatorname{Hom}_{kG}(M,kG)$ are isomorphic as right $kG$-modules. An explicit isomorphism $M^\ast\to M^\vee$ is given by

$$\phi\mapsto\left(m \mapsto \sum_{g\in G} \phi(g^{-1}m)g\right).$$

The inverse isomorphism is simply picking the coefficient of the unit element of $G$.

For Clifford algebras things should go the same way, at least over $\mathbb{R}$ and $\mathbb{C}$. Let $(e_1,\dotsc, e_n)$ be an orthonormal basis of $V$, and let $G$ be the subgroup of $Pin(V)$ consisting of the elements $\{\pm e_1^{\epsilon_1}\dotsm e_n^{\epsilon_n}\}$ where $\epsilon_i\in \{0,1\}$. Then $\Cl(V)$-modules are equivalently $G$-modules such that the element $-1$ of $G$ acts as the multiplication by $-1$. One can then define an isomorphism $M^\ast\to M^\vee$ by

$$\phi\mapsto\left(m \mapsto \frac{1}{2}\sum_{g\in G} \phi(g^{-1}m)g\right)=\left(m \mapsto \sum_{g\in G^+} \phi(g^{-1}m)g\right),$$

where $G^+\subset G$ is the subset $\{e_1^{\epsilon_1}\dotsm e_n^{\epsilon_n}\}$ of “positive” elements. Here the sums are done in $\Cl(V)$. To write the inverse isomorphism one notices that there exist a linear morphism $\pi\colon \Cl(V)\to k$ that sends $1$ to $1$ and all other elements of $G^+$ to $0$.

Now the questions (assuming all of the above is correct), are:

i) can one give the isomorphism of left Clifford modules $M^\ast\cong M^\vee$ in an intrinsic way, without using an orthonormal basis (this would possibly give an isomorphism over an arbitrary field $k$)?

ii) if the intrinsic approach fails, can one at least show that the isomorphism above is independent of the chosen orthonormal basis? (This should be a simple check, but I'm postponing it in the hope of a positive answer to i).)

Let $Cl(V)$ be the Clifford algebra of a quadratic vector space $(V,Q)$ over $k$, and let $M$ be a (left) $Cl(V)$-module. The dual Clifford module is typically defined as the linear dual $M^\ast=\mathrm{Hom}_k(M,k)$ with the Clifford action given by defining $g\cdot \phi$ as $g\cdot \phi\colon m\mapsto \phi(g^{\perp}m)$, where $\perp$ is the standard antinvolution of $Cl(V)$ given by $(v_1v_2\cdots v_n)^\perp=v_n\cdots v_2v_1$. This is similar (not to say it is the same) as for the dual represetation of a finite group $G$, where the standard antinvolution of $kG$ is $g\mapsto g^{-1}$.

Yet there is another, a priori distinct and actually more natural, definition of a dual Clifford module: one can consider ${M^\vee}=\mathrm{Hom}_{Cl(V)}(M,Cl(V))$ with its natural right $Cl(V)$-module structure.

This is equivalently a left $Cl(V)^{opp}$-module structure, and since $\perp$ is an algebra isomorphism from $Cl(V)$ to $Cl(V)^{opp}$, we see that $M^\vee=\mathrm{Hom}_{Cl(V)}(M,Cl(V))$ carries a left $Cl(V)$-module structure via $\perp$.

Now, one may wonder whether $M^\ast$ and $M^\vee$ are isomorphic as left $Cl(V)$-modules. Equivalently, and more naturally, one is wondering whether $M^\ast$ and $M^\vee$ are isomorphic as right $Cl(V)$-modules, where the right $Cl(V)$-module structure on $M^\ast$ is given by $\phi\cdot g\colon m\mapsto \phi(gm)$.

For the representations of a finite group $G$ this is precisely what happens: $M^\ast$ and $M^\vee=\mathrm{Hom}_{kG}(M,kG)$ are isomorphic as right $kG$-modules. An explicit isomorphism $M^\ast\to M^\vee$ is given by

$\phi\mapsto\left(m \mapsto \sum_{g\in G} \phi(g^{-1}m)g\right)$

The inverse isomorphism is simply picking the coefficient of the unit element of $G$.

For Clifford algebras things should go the same way, at least over $\mathbb{R}$ and $\mathbb{C}$. Let $(e_1,\dots, e_n)$ be an orthonormal basis of $V$, and let $G$ be the subgroup of $Pin(V)$ consisting of the elements $\{\pm e_1^{\epsilon_1}\cdots e_n^{\epsilon_n}\}$ where $\epsilon_i\in \{0,1\}$. Then $Cl(V)$-modules are equivalently $G$-modules such that the element $-1$ of $G$ acts as the multiplication by $-1$. One can then define an isomorphism $M^\ast\to M^\vee$ by

$\phi\mapsto\left(m \mapsto \frac{1}{2}\sum_{g\in G} \phi(g^{-1}m)g\right)=\left(m \mapsto \sum_{g\in G^+} \phi(g^{-1}m)g\right)$,

where $G^+\subset G$ is the subset $\{e_1^{\epsilon_1}\cdots e_n^{\epsilon_n}\}$ of "positive" elements. Here the sums are done in $Cl(V)$. To write the inverse isomorphism one notices that there exist a linear morphism $\pi\colon Cl(V)\to k$ that sends $1$ to $1$ and all other elements of $G^+$ to $0$.

Now the questions (assuming all of the above is correct), are:

i) can one give the isomorphism of left Clifford modules $M^\ast\cong M^\vee$ in an intrinsic way, without using an orthonormal basis (this would possibly give an isomorphism over an arbitrary field $k$)

ii) if the intrinsic approach fails, can at least one show that the isomorphism above is independent of the chosen orthonormal basis? (this should be a simple check, but I'm postponing it in the hope of a positive answer to i)).

$\DeclareMathOperator\Cl{Cl}$Let $\Cl(V)$ be the Clifford algebra of a quadratic vector space $(V,Q)$ over $k$, and let $M$ be a (left) $\Cl(V)$-module. The dual Clifford module is typically defined as the linear dual $M^\ast=\operatorname{Hom}_k(M,k)$ with the Clifford action given by defining $g\cdot \phi$ as $g\cdot \phi\colon m\mapsto \phi(g^{\perp}m)$, where $\perp$ is the standard antiinvolution of $\Cl(V)$ given by $(v_1v_2\dotsm v_n)^\perp=v_n\dotsm v_2v_1$. This is similar (not to say it is the same) to the dual represetation of a finite group $G$, where the standard antiinvolution of $kG$ is $g\mapsto g^{-1}$.

Yet there is another, a priori distinct and actually more natural, definition of a dual Clifford module: one can consider ${M^\vee}=\operatorname{Hom}_{\Cl(V)}(M,\Cl(V))$ with its natural right $\Cl(V)$-module structure.

This is equivalently a left $\Cl(V)^\text{opp}$-module structure, and since $\perp$ is an algebra isomorphism from $\Cl(V)$ to $\Cl(V)^\text{opp}$, we see that $M^\vee=\operatorname{Hom}_{\Cl(V)}(M,\Cl(V))$ carries a left $\Cl(V)$-module structure via $\perp$.

Now, one may wonder whether $M^\ast$ and $M^\vee$ are isomorphic as left $\Cl(V)$-modules. Equivalently, and more naturally, one is wondering whether $M^\ast$ and $M^\vee$ are isomorphic as right $\Cl(V)$-modules, where the right $\Cl(V)$-module structure on $M^\ast$ is given by $\phi\cdot g\colon m\mapsto \phi(gm)$.

For the representations of a finite group $G$ this is precisely what happens: $M^\ast$ and $M^\vee=\operatorname{Hom}_{kG}(M,kG)$ are isomorphic as right $kG$-modules. An explicit isomorphism $M^\ast\to M^\vee$ is given by

$$\phi\mapsto\left(m \mapsto \sum_{g\in G} \phi(g^{-1}m)g\right).$$

The inverse isomorphism is simply picking the coefficient of the unit element of $G$.

For Clifford algebras things should go the same way, at least over $\mathbb{R}$ and $\mathbb{C}$. Let $(e_1,\dotsc, e_n)$ be an orthonormal basis of $V$, and let $G$ be the subgroup of $Pin(V)$ consisting of the elements $\{\pm e_1^{\epsilon_1}\dotsm e_n^{\epsilon_n}\}$ where $\epsilon_i\in \{0,1\}$. Then $\Cl(V)$-modules are equivalently $G$-modules such that the element $-1$ of $G$ acts as the multiplication by $-1$. One can then define an isomorphism $M^\ast\to M^\vee$ by

$$\phi\mapsto\left(m \mapsto \frac{1}{2}\sum_{g\in G} \phi(g^{-1}m)g\right)=\left(m \mapsto \sum_{g\in G^+} \phi(g^{-1}m)g\right),$$

where $G^+\subset G$ is the subset $\{e_1^{\epsilon_1}\dotsm e_n^{\epsilon_n}\}$ of “positive” elements. Here the sums are done in $\Cl(V)$. To write the inverse isomorphism one notices that there exist a linear morphism $\pi\colon \Cl(V)\to k$ that sends $1$ to $1$ and all other elements of $G^+$ to $0$.

Now the questions (assuming all of the above is correct), are:

i) can one give the isomorphism of left Clifford modules $M^\ast\cong M^\vee$ in an intrinsic way, without using an orthonormal basis (this would possibly give an isomorphism over an arbitrary field $k$)?

ii) if the intrinsic approach fails, can one at least show that the isomorphism above is independent of the chosen orthonormal basis? (This should be a simple check, but I'm postponing it in the hope of a positive answer to i).)