Mapping tensor products into a Clifford algebra

[jv07cs]

Considering a vector space $W = V\oplus V^*$ equipped with quadratic form Q such that we have a clifford algebra $Cl(W, Q)$. How can I map elements of $V\otimes V^*$ into elements of $Cl(W, Q)$? What about elements of $V^* \otimes V$, $V\otimes V$ and $V^* \otimes V^*$ into $Cl(W, Q)$?

 

[fresh_42]

What is your desire?
What is your favorite color?
How fast is a swallow?

Means, what are you aiming to achieve? I like to think of $V\otimes V^*$ as an element of $\operatorname{End}(V)$ given by $\left(\sum_\rho u_\rho \otimes \overline{v}_\rho\right)(w)=\sum_\rho \overline{v}_\rho(w)\cdot u_\rho\,,$ i.e. a matrix. Now you are asking how you can map an endomorphism of $V$ into $Cl(W,Q).$ This looks a bit arbitrary to me.

Let's see whether the realization of $Cl(W,Q)$ as a tensor algebra can help. This would mean to map
$$
V\otimes V^* \stackrel{\varphi}{\longrightarrow} \bigoplus_{n\ge 0}(V\oplus V^*)^{\otimes n} /\underbrace{\bigl\langle (u+\overline{v})+(\overline{v}+u)+Q(u,\overline{v}) \bigr\rangle }_{=\mathcal{I}}.
$$
This view has two advantages: The ideal $\mathcal{I}$ should be invariant under $\varphi$, and we can see the arbitrariness better since $V\otimes V^*$ has many occurrences on the right and you can simply choose one of them.

Another method would be the following:
Consider $V$ and $V^*$ as elements of $Cl(W,Q)$ via the embeddings $u\longmapsto (u+0) \cdot 1_{Cl}$ and $\overline{v}\longmapsto (0+\overline{v})\cdot 1_{Cl}$ and the tensor product as multiplication in $Cl(W,Q).$ Then
$$
\varphi \left(\sum_\rho u_\rho\otimes \overline{v}_\rho\right)=\sum_\rho \left( \left(u_\rho+0) \cdot 1_{Cl}\right)\cdot \left((0+\overline{v}_\rho)\cdot 1_{Cl}\right)\right)
$$
In this case, the arbitrariness is hidden in the embedding.

My original question stands: what is your desire?

 

[jv07cs]

What is your desire?
What is your favorite color?
How fast is a swallow?

Means, what are you aiming to achieve? I like to think of $V\otimes V^*$ as an element of $\operatorname{End}(V)$ given by $\left(\sum_\rho u_\rho \otimes \overline{v}_\rho\right)(w)=\sum_\rho \overline{v}_\rho(w)\cdot u_\rho\,,$ i.e. a matrix. Now you are asking how you can map an endomorphism of $V$ into $Cl(W,Q).$ This looks a bit arbitrary to me.

Let's see whether the realization of $Cl(W,Q)$ as a tensor algebra can help. This would mean to map
$$
V\otimes V^* \stackrel{\varphi}{\longrightarrow} \bigoplus_{n\ge 0}(V\oplus V^*)^{\otimes n} /\underbrace{\bigl\langle (u+\overline{v})+(\overline{v}+u)+Q(u,\overline{v}) \bigr\rangle }_{=\mathcal{I}}.
$$
This view has two advantages: The ideal $\mathcal{I}$ should be invariant under $\varphi$, and we can see the arbitrariness better since $V\otimes V^*$ has many occurrences on the right and you can simply choose one of them.

Another method would be the following:
Consider $V$ and $V^*$ as elements of $Cl(W,Q)$ via the embeddings $u\longmapsto (u+0) \cdot 1_{Cl}$ and $\overline{v}\longmapsto (0+\overline{v})\cdot 1_{Cl}$ and the tensor product as multiplication in $Cl(W,Q).$ Then
$$
\varphi \left(\sum_\rho u_\rho\otimes \overline{v}_\rho\right)=\sum_\rho \left( \left(u_\rho+0) \cdot 1_{Cl}\right)\cdot \left((0+\overline{v}_\rho)\cdot 1_{Cl}\right)\right)
$$
In this case, the arbitrariness is hidden in the embedding.

My original question stands: what is your desire?

My original question was very unspecific indeed. I want to investigate endomorphisms inside $Cl(W, Q)$ induced by endomorphisms in $V$ and $V^*$.

Taking $T = T_i^j e_j \otimes e^i \in End(V)$, for example. I want to view this $T$ as an element of $Cl(W, Q)$, let's denote it by $\phi(T)$ (where $\phi$ is the clifford map), so that I can study $\phi(Tv)$ (where $Tv \equiv (Tv, 0) \in W$) as some relation between $\phi(T)$ and $\phi(v)$.

From your reply, if I understood correctly, there is an arbitrariness in the choice of $\phi(T)$. My question is what would I have to verify to ensure that my choice of $\phi(T)$ is valid? How could I verify if, for example, $\phi(T) = T_i^j \phi(e_j)\phi(e^i)$ and $\phi(S) = T^j_i\phi(e^i)\phi(e_j)$ (where $S = T^j_i e^i\otimes e_j$, notice the change in the ordering of the tensor product) are valid choices?

 

[fresh_42]

My original question was very unspecific indeed. I want to investigate endomorphisms inside $Cl(W, Q)$ induced by endomorphisms in $V$ and $V^*$.

Taking $T = T_i^j e_j \otimes e^i \in End(V)$, for example. I want to view this $T$ as an element of $Cl(W, Q)$, let's denote it by $\phi(T)$ (where $\phi$ is the clifford map), so that I can study $\phi(Tv)$ (where $Tv \equiv (Tv, 0) \in W$) as some relation between $\phi(T)$ and $\phi(v)$.

What do you mean by Clifford map? The quotient of the tensor algebra by the ideal $\mathcal{I}$ I mentioned, or your proposed map? I don't think the ordering plays a role as long as $V$ and $V^*$ are distinguishable. I wouldn't use $V\cong V^*$ anywhere as it creates a new kind of arbitrariness since this isomorphism isn't natural.

From your reply, if I understood correctly, there is an arbitrariness in the choice of $\phi(T)$. My question is what would I have to verify to ensure that my choice of $\phi(T)$ is valid? How could I verify if, for example, $\phi(T) = T_i^j \phi(e_j)\phi(e^i)$ and $\phi(S) = T^j_i\phi(e^i)\phi(e_j)$ (where $S = T^j_i e^i\otimes e_j$, notice the change in the ordering of the tensor product) are valid choices?

Bilinearity is usually achieved per definition: you define the map on dyads $u\otimes \overline{v}$ and extend it bilinearly. You have a tiny part $V\otimes V^*$ of the tensor algebra of $W.$ You can embed it (in many ways) and build the quotient by $\mathcal{I}$ afterward.

Since $V$ and $V^*$ are different, you won't have a diagonal $v\otimes v$ that had to be mapped to $-Q(v)\cdot 1_{Cl}.$ You can simply use the multiplication in the Clifford algebra:
$$
u\otimes \overline{v}\longmapsto (u+0) \cdot_{Cl} (0+\overline{v})\,.
$$

 

[jv07cs]

What do you mean by Clifford map? The quotient of the tensor algebra by the ideal I I mentioned, or your proposed map?

I am referring to the following definition of a clifford algebra:

But I believe it is equivalent to the quotient algebra definition. My understanding of clifford algebras is quite basic, I hadn't really studied its definition as a quotient algebra, but, if I understood correctly, the equivalence classes satisfy $[w\otimes w] = [Q(w)] \in Cl(W,Q) = T(V)/\mathcal{I}$, then I represent this as

$$w\cdot_{Cl} w = Q(w)\cdot_{Cl} 1_{Cl}$$

Or, equivalently (I think), using the clifford mapping $\phi$

$$\phi(w)\cdot_{Cl} \phi(w) = Q(w)\cdot_{Cl} 1_{Cl}$$

Is this correct?

I wouldn't use V≅V∗ anywhere as it creates a new kind of arbitrariness since this isomorphism isn't natural.

Well, I could define a natural isomorphims by introducing a metric structure on $V$. Don't know if that would make the ordering significant.

Bilinearity is usually achieved per definition: you define the map on dyads u⊗v― and extend it bilinearly. You have a tiny part V⊗V∗ of the tensor algebra of W. You can embed it (in many ways) and build the quotient by I afterward.

Would it then suffice to simply define a valid homomorphism between $End(V)$ and $Cl(W, Q)$?

 

[fresh_42]

Is this correct?

Yes, except that it should be $w\cdot_{Cl} w=-Q(w)\cdot 1_{Cl}.$ We immitate the imaginary unit!

Well, I could define a natural isomorphims by introducing a metric structure on $V$. Don't know if that would make the ordering significant.

I think that distinction is important, not so much the order. E.g., if we consider mappings $V\otimes V \rightarrow Cl(V,Q)\subseteq Cl(W,Q)$ then $v\otimes v$ must map to $-Q(v)\cdot 1_{Cl}$ which may restrict the possible images for $v,$ i.e. we have another condition that has to be fulfilled. This is probably no problem but we have to keep it in mind. We map the diagonal on $\mathbb{R}\cdot 1_{Cl}.$ There is no correspondence in a domain $V\otimes V^*.$

Would it then suffice to simply define a valid homomorphism between $End(V)$ and $Cl(W, Q)$?

That is my understanding, yes.

 

[jv07cs]

Yes, except that it should be $w\cdot_{Cl} w=-Q(w)\cdot 1_{Cl}.$ We immitate the imaginary unit!

I think that distinction is important, not so much the order. E.g., if we consider mappings $V\otimes V \rightarrow Cl(V,Q)\subseteq Cl(W,Q)$ then $v\otimes v$ must map to $-Q(v)\cdot 1_{Cl}$ which may restrict the possible images for $v,$ i.e. we have another condition that has to be fulfilled. This is probably no problem but we have to keep it in mind. We map the diagonal on $\mathbb{R}\cdot 1_{Cl}.$ There is no correspondence in a domain $V\otimes V^*.$


That is my understanding, yes.

Got it. Just one last quick question: If I consider $B$ to be the bilinear form associated with the quadratic form $Q$ (so $Q(w) := B(w,w)$). The ideal could be expressed as

$$\mathcal{I} = \langle w\otimes w' + w'\otimes w - 2B(w,w')\rangle$$

And $e_j\otimes e^i$ would be equivalent to $2B(e_j, e^i) - e^i\otimes e_j$, so

$$[e_j\otimes e^i] = [2B(e_j, e^i) - e^i\otimes e_j] \implies e_j\cdot_{Cl} e^i = 2B(e_j, e^i)\cdot_{Cl} 1_{Cl} - e^i\cdot_{Cl} e_j$$

which is the fundamental defining relation of the clifford algebra with $w = e_j$ and $w' = e^i$. Would it correct to say that this implies that $T = T_i^j e_j\otimes e^i$ is mapped to $T_i^j e_j\cdot_{Cl} e^i$?

 

[fresh_42]

Got it. Just one last quick question: If I consider $B$ to be the bilinear form associated with the quadratic form $Q$ (so $Q(w) := B(w,w)$). The ideal could be expressed as

$$\mathcal{I} = \langle w\otimes w' + w'\otimes w - 2B(w,w')\rangle$$

I haven't checked. My book uses $Q(w+w')=Q(w)+Q(w')+2B(w,w').$ Unfortunately, it doesn't mention the definition as a quotient of a tensor algebra and I'm not certain that Wikipedia can be trusted.

If we assume, that we can trust it, then the ideal is generated by $w\otimes w - Q(w)1.$ Let's see.
\begin{align*}
(w+w') \otimes (w'+w) - Q(w+w')1&=w\otimes w' +w'\otimes w +Q(w)1+Q(w')1-Q(w+w')1\\&=w\otimes w' +w'\otimes w -2B(w,w')
\end{align*}
(Sorry for thinking loud.)

And $e_j\otimes e^i$ would be equivalent to $2B(e_j, e^i) - e^i\otimes e_j$, so

$$[e_j\otimes e^i] = [2B(e_j, e^i) - e^i\otimes e_j] \implies e_j\cdot_{Cl} e^i = 2B(e_j, e^i)\cdot_{Cl} 1_{Cl} - e^i\cdot_{Cl} e_j$$

which is the fundamental defining relation of the clifford algebra with $w = e_j$ and $w' = e^i$. Would it correct to say that this implies that $T = T_i^j e_j\otimes e^i$ is mapped to $T_i^j e_j\cdot_{Cl} e^i$?

Not sure I can follow your calculations here, but I'd answer this question with a yes. That is what I would try first, the quotient map from the tensor algebra:
$$
V\otimes V^* \subseteq \mathcal{T}(W) \twoheadrightarrow Cl(W,Q)
$$
O'Meara has an interesting definition of a Clifford algebra which might be of interest here:

We call $Cl(W,Q)$ a Clifford algebra over the quadratic space $(W,Q)$ if it satisfies the following universal mapping property: given any algebra $A$ compatible with $W,$ i.e. $W\subseteq A$ and $w^2=Q(w)\cdot 1_A$, there is exactly one algebra homomorphism $\varphi\, : \,Cl(W,Q)\longrightarrow A$ such that $\varphi(w)=w$ for all $w\in W.$

Unfortunately, $V\otimes V^* \not\subseteq W=V\oplus V^*$ and you are trying to find a homomorphism from $A=\operatorname{End}(V)$ into $Cl(W,Q)$ instead of the other way around. This contrast to the universal property creates the arbitrariness I have spoken about. That's why I would take the detour by the tensor algebra $\mathcal{T}(W)=\mathcal{T}(V\oplus V^*).$ But there is more than one way to embed $\operatorname{End}(V) \hookrightarrow \mathcal{T}(W).$