Quaternions in Clifford algebras

[mnb96]

Hello,
it is known quaternions are isomorphic to $$\mathcal{C}\ell^{+}_{3,0}$$, which is the even subalgebra of $$\mathcal{C}\ell_{3,0}$$

Is it possible to find an isomorphism between $$\mathcal{C}\ell_{2,0}$$ and $$\mathbb{H} \cong \mathcal{C}\ell^{+}_{3,0}$$ ?

*** my attempt was: ***

Let's consider $$\{1,e_1, e_2, e_{12}\}$$ and the morphism f defined as follows:

$$f(1)=1$$

$$f(e_{32})=e_1$$

$$f(e_{13})=e_2$$

$$f(e_{21})=e_{21}$$

This almost works, in fact:
$f(xy)=f(x)f(y)$ always holds with one exception:

$$f(e_1 e_2) = -e_{21} = f(e_2)f(e_1)$$
So, in this case, the morphism f does not preserve well the geometric product.
Is it possible to make the whole thing work?

 

[Simon_Tyler]

What you have written doesn't quite make sense. Since you have not defined $$f(e_i)$$ for $$i=1,2$$.

I think what you wanted was
$$f(e_{12}) = -f(e_{21}) = -e_{21} = -e_2 e_1 = -f(e_{13})f(e_{32}) =-f(e_{13}e_{32})=f(e_{12})$$
which is all as it should be.

 

[arkajad]

You got all right squares and all right anticommutators. So it works.
And indeed , there is something wrong with your last formula. It is not clear what you were trying to do.

 

[mnb96]

thanks for both your replies.
Maybe I now understood the mistake.

Now, please tell me if this is the correct way to proceed:
I want to find an isomorphism between the subalgebra of quaternions $$\mathcal{C}\ell^{+}_{3,0}$$ and the algebra $$\mathcal{C}\ell_{2,0}$$

I define $$f:\mathcal{C}\ell^{1}_{3,0} \rightarrow \mathcal{C}\ell^{1}_{2,0}$$ as:

$$f(1)=1$$
$$f(e_1)=e_1$$
$$f(e_2)=e_2$$
$$f(e_3)=e_{12}$$

Now, I simply verify that the quaternion rules $$i^2 = j^2 = k^2 = ijk = -1$$, where $i=e_{32}$ $j=e_{13}$ $k=e_{21}$ correctly hold as follows $$f(i^2) = f(j^2) = f(k^2) = f(ijk) = f(-1)$$

Does this make more sense?

 

[arkajad]

Well, you need to show that $$f(e_i e_j)=f(e_i)f(e_j)$$.
If you have that $$f(e_1 e_2 e_3)=f(e_1)f(e_2)f(e_3)$$ then $$f(e_1e_2)=f(e_3)$$
$$f(e_1)f(e_2)f(e_3)=-1,$$ therefore (multiply on the right by $$f(e_3)$$)
$$f(e_1)f(e_2)=f(e_3)=f(e_1 e_2),$$
and you are home.
If this is what you mean, then you are alright.

 

[arkajad]

What you did is a particular case of a general theorem:

If the quadratic form $$q$$ on $$E$$ is non-zero, the even subalgebra $$C_+$$ of $$C(E;q)$$ can be, in a natural way, considered as the Clifford algebra $$C(E_1;q_1)$$ of the subspace $$E_1=e_1^{\perp}$$
of $$E$$, orthogonal to a regular vector $$e_1$$, where $$q_1=-q(e_1)q$$.

For this you define $$f(y)=e_1 y,$$ $$y\in e_1^{\perp}$$.

Then $$f(y)^2 = e_1 y e_1 y = -e_1^2 y^2 = -q(e_1)q(y).$$

P.S. I am using the convention $$C(E;q)= \mathcal{C} l (E,-q)$$

 

[mnb96]

Sorry,
I am a bit confused again now.
After sleeping over it, What I wrote in my previous post doesn't sound correct anymore, because the quaternion units do not square to -1 in $$\mathcal{C}\ell_{2,0}$$.

For example, let's take one quaternion unit in $$\mathcal{C}\ell_{3,0}$$:

$$i = e_{32}$$

We have that:

$$f(e_{32}e_{32}) = f(-1) = -1 \neq f(e_{32})f(e_{32})$$

so f is not a morphism because it does not preserve the geometric-product.PS: @arkajad: could you please give me a reference for that general theorem (if possible)?

 

[arkajad]

Rene Deheuvels, "Tenseurs et Spineurs", Presses Universitaires de France, (1993), pp. 247-248 (Very good book. Good to have in the library.)

I have scanned for you the Theorem, its proof, and and example

 

[mnb96]

... it seems to me that it is not possible to create an isomorphism between $$\mathcal{C}\ell_{2,0}$$ and the quaternion subalgebra $$\mathcal{C}\ell^{+}_{3,0}$$, unless one reverses the metric signature into:

$$\mathcal{C}\ell_{0,2}$$

This is probably the only way to obtain $$e_i e_i = -1$$.

 

[arkajad]

Which convention are you using for Cl(q)?

xy+yx = 2(x,y) or xy+yx = - 2(x,y)

I should have asked this question first.

 

[mnb96]

I was assuming the first one: xy+yx = 2(x,y).
Sorry for having skipped that.

 

[arkajad]

Then indeed you have an isomorphism of $$Cl^+(3,0)$$ with $$Cl(0,2)$$

P.S. I was too quick in my first replies!