Proving No Non-Trivial Ideals in Clifford-t Algebra for t ≠ 0 | V = W ⊕ W*

[AKG]

Let $\mathbb{K}$ be a field. Assume that any vector spaces mentioned hereafter have $\mathbb{K}$ underlying them. W is a vector space, and W* is the dual space. If Y, Z are vector spaces, define the vector space:

Y * Z = Span{F(y,z) | y in Y, z in Z}

So if (y,z) and (y',z') are distinct, then F(y,z) and F(y',z') are considered to be linearly independent in Y * Z. Define:

U = Span{F(0,z), F(y,0), F(ay,z) - aF(y,z), F(y,az) - aF(y,z), F(y + w,z) - [F(y,z) + F(w,z)], F(y,z + x) - [F(y,z) + F(y,x)] | a is in K; y, w are in Y; z, x are in Z}

Define the vector space:

$$Y \otimes Z = (Y*Z)/U$$

and define, for all y in Y and all z in Z

$$y \otimes z = F(y,z) + U$$

i.e. the coset of U containing F(y,z)

If V is a vector space, define T(V) to be the graded algebra:

$$\mathbb{K} \oplus V \oplus (V \otimes V) \oplus (V \otimes V \otimes V) \oplus \dots$$

Now let $V = W \oplus W^*$. Define I(V)_t^odd to be the (2-sided) ideal of T(V) generated by all elements of the form:

$$(w_1,f_1) \otimes (w_2,f_2) + (w_2,f_2)\otimes (w_1,f_1) - t(f_1(w_1) + f_2(w_1))$$

where t is some element of our underlying field. The (2-sided) ideal generated by those elements is the smallest subspace of T(V) (i.e. it is a vector space) containing all those elements and also satisfying the condition that for every element a of T(V) and every element i of the ideal, ai and ia are in the ideal. Finally, define the Clifford-t algebra:

$$Cl_t(V) = T(V)/I(V)_t ^{odd}$$

where, recall, t is an element of the underlying field $\mathbb{K}$ and V is $W \oplus W^*$, where W is some $\mathbb{K}$-vector space.

The problem I'm assigned to do is to show that the Clifford-t algebra defined above has no non-trivial (2-sided) ideals for t not equal to 0. So I think I want to show that if I have some element of this C = Cl_t(V) in my ideal that is non-zero, then I can force that ideal to have the unit of C, and thus every element of C. Let I = I(V)_t^odd. The unit element of C is 1 + I, where 1 is the unit element of T(V). The unit element of T(V) is just the multiplicative identity of $\mathbb{K}$. So I think I want to say that if x is some non-zero element of T(V) - I, then the vector space J containing x+y for each y in I and such that for all j in J and a in T(v), aj and ja are in J is such that there is some x' in J and y' in I such that x' - y' = 1.

So I start with:

{x}

Then I create:

{x + i | i in I}

Then I create:

{a(x + i), (x + i)a, a(x + i)b | a,b in T(V), i in I}

Then I make it a vector space (close it under addition and scalar multiplication):

Span{a(x + i), (x + i)a, a(x + i)b | a,b in T(V), i in I}

So I have to show that for any non-zero x in T(V), I can pick (for some n,m) n+m+p scalars k_i, n+m+2p elements of T(V) a_i, and n+m+p+1 elements of the ideal j₁, ..., j_n, y such that:

$$\sum _{i=1} ^n k_ia_i(x + j_i) + \sum _{i=1} ^m k_i(x + j_i)a_i + \sum _{i=1} ^p k_ia_i(x + j_i)a_{p+i} - y = 1$$

Am I on any sort of a right track? What do I do from here? Is there a good way of thinking about the elements of I?

EDIT: Actually, since the identity element is an element of T(V), I just need to look at the sum:

$$\sum _{i=1} ^n k_ia_i(x + j_i)x_{n+i} - y = 1$$

i.e. things of the form a(x+j) are just of the form a(x+j)1 which is of the form a(x+j)b. So to restate, if x is an element of T(V) that is not an element of I, I need to show that for some n, I can pick n scalars {k₁, ..., k_n}, 2n elements of T(V) {a<sub>1, ..., a2n}, and n+1 elements of I {j1, ..., jn, y} such that

k1a1(x + j1)an+1 + k2a2(x + j2)an+2 + ... + knan(x + jn)a2n - y = 1

This is formally what I want to do. However, I have no intuitive sense of what's going on, or what these elements x and j, etc. are "like" so I don't know what to do from here. I had a previous question that asked to show that Hom(V,V) where V is a finite-dimensional vector space has no non-trivial 2-sided ideals. I did this by treating Hom(V,V) as the set of nxn matrices, and saying that if I have a non-zero matrix, I can row reduce it, then left and right multiply it by the matrix that has a 1 in the first entry, and zeroes every where else, then multiply it by a particular scalar so I'm left with a matrix that just has a 1 in the top-left and zeroes everywhere else. Then I can make n-1 more matrices (where dim(V) = n) that have zeroes everywhere except a 1 in the second, third, ..., or nth place, by elementary row operations. I can add these n matrices together to get the identity. Then I can get everything in Hom(V,V) so this ideal is trivial. In this case, I was able to think of Hom(V,V) as something more intuitive or easier to deal with (matrices) and was able to think of things I could do to it (row reduce, apply elementary row operations to get the "1" to move down the diagonal, etc.). I have no familiarity with the structure being dealt with in the question of this thread, so I don't know how to go about doing the problem. Is C = Clt(V) isomorphic to something nice like an algebra of matrices or something like that?</sub>

 

[AKG]

Looking at it some more, I don't even need the scalars k_i or the n elements of the ideal, j_i. So all I need to do is show that I can choose 2n elements of T(V) and one element of I such that:

a₁xa_n+1 + ... + a_nxa_2n - y = 1

If I can get the left side to equal some non-zero k of K, then multiplying the left side by 1/k gives me what I want (since (1/k)y is still an element of the ideal, and k(a_ixa_i+n) = (ka_i)xa_i+n), so it suffices to show that I can make some expression like:

a₁xa_n+1 + ... + a_nxa_2n

differ from some element y of I by k, where k is a non-zero element of the field. Well I know that t is a non-zero element of the field, so maybe (and I couldn't say why) I should aim to show that I can get that expression to differ from some y in I by t.

Any hints or suggestions as to how to approach this problem? What is the typical element x of T(V) - I? What is the typical element of I like (other than some ugly linear combination?).

I know that the construction of the tensor product of two vector spaces, although it looks like the quotient V*W by U, where U appears to be some vector space spanned by some complicated basis, is just making F bilinear. Or in other words, it just defines a bilinear multiplication for vectors, but sends the product of two vectors into a new space. Even though U is a strange looking set, modding out by U does something that is not strange, it is something that is relatively easy to grasp. Now this ideal I is strange to me. But is modding out T(V) by this ideal something easy to grasp? Is there a nice way to think about the quotient T(V)/I aside from thinking of it as the set of cosets of I, where I is that strange ideal?

 

[Hurkyl]

So, your question is: what is this messy thing I've created? (At least one of them is)

Let me start with a notational simplification: you have a canonical embedding of W and W* in W+W*, and IMHO it is much simpler to understand if you write things in terms of this embedding. I.E. in W+W*, we can write (w, f) as w + f.

This suggests an obvious manipulation of the defining property of elements of I -- I'll give you a chance to work that out, and see where that goes.

Also, it might be useful to see if you can identify especially simple elements of I.

(By the way, that is a typo in the defining property of I, right?)


More simplification: in an associative algebra, it just seems easier to me to write multiplication as xy, rather than the $x \otimes y$ that you're using.


Also, it may help to think about different interpretations of doing arithmetic in a quotient object.

 

[AKG]

Yes, it is, unfortunately I can't edit it now. Using juxtaposition to denote the product of two elements instead of the tensor symbol, and letting w denote the element (w,0) of W + W*, letting f denote (0,f), the defining elements of I take the form:

w₁w₂ + w₁f₂ + f₁w₂ + f₁f₂ + w₂w₁ + w₂f₁ + f₂w₁ + f₂f₁ + t(f₁(w₁) + f₂(w₂))

For some simple elements of the ideal, we get every element of the form fg + gf, and wv + vw. Combining this with what the defining elements look like, we know that I contains elements of the form:

w₁f₂ + f₁w₂ + w₂f₁ + f₂w₁ + t(f₁(w₁) + f₂(w₂))

Actually, using reasoning similar to what I used to get fg + gf to be contained in I, I can show that elements of the form wf + fw are in I, so I contains elements of the form:

t(f₁(w₁) + f₂(w₂))

Arguably, I contains the entire field K.

Going back to:

a₁xa_n+1 + ... + a_nxa_2n - y = 1

I can just let y be -1, let n = 1, let a₁ = a_n+1 = 0 and the equation is satisfied. I'll have to write this out formally later on to convince myself that I haven't cheated anywhere. Would you be able to tell at a glance whether I have or not?

Just to clarify, I know that, for example, I contains elements of the form fw + wf because if we look at:

w₁w₂ + w₁f₂ + f₁w₂ + f₁f₂ + w₂w₁ + w₂f₁ + f₂w₁ + f₂f₁ + t(f₁(w₁) + f₂(w₂))

and let w₂ and f₁ be the zero vector and zero functional respectively, we get:

w₁w₂ + w₁f₂ + f₁w₂ + f₁f₂ + w₂w₁ + w₂f₁ + f₂w₁ + f₂f₁ + t(f₁(w₁) + f₂(w₂)) = w₁0 + w₁f₂ + 00 + 0f₂ + 0w₁ + 00 + f₂w₁ + f₂0 + t(0(w₁) + f₂(0)) = w₁f₂ + f₂w₁

I also know that I contains the entire field because if I contains elements of the big ugly form, plus elements of the form wv + vw, fg + gf, and fw + wf, then since it's a vector space closed under addition, I can subtract elements of the above three small forms from the big ugly form and just get t(f(w) + g(v)). So if it contains all elements of the form t(f(w) + g(v)) then it does so for v = 0, f = u*, and w = ku for some scalar k, so we get all elements of the form:

t(u*(ku)) = kt(u*(u)) = kt

i.e. u* is the functional dual to u, it sends u to the identity of the field, 1. If I have all elements of the form kt, since t is non-zero, I have every element of the field. How does all this sound?

Thanks a lot for the help.

 

[AKG]

I definitely cheated somewhere. If the ideal contains every element of the field, then it contains the unit element, and so it contains every element of T(V). This would make our clifford algebra C simply T(V)/T(V) = {T(V)}, and the whole exercise would be trivial. I doubt it's meant to be so, so I guess I've got to look over this again. Was anything in the above post on the right track?

 

[Hurkyl]

Hrm. You sure you have the definition right? My scribblings suggest that letting I be the elements of the form:

$$(w_1+f_1)(w_2+f_2) + (w_2+f_2)(w_1+f_1) - t(f_1(w_2) + f_2(w_1))$$

gives you something fairly natural.

But with what you've given, I think you're right, since in

$$w_1f_2 + f_1w_2 + w_2f_1 + f_2w_1 + t(f_1(w_1) + f_2(w_2))$$

if I let $w_2 = 0$, we get:

$$w_1f_2 + f_2w_1 + t f_1(w_1)$$

Letting $f_2 = 0$, we have that I contains $t f_1(w_1)$ for any $f_1$ and $w_1$.


(Anyways, yes, you do seem to be using a good approach)

 

[AKG]

Yes, $\mathcal{I}$ should be elements of that form. There was a typo in my original definition, but I "corrected" the wrong part of it. So we have:

$$(w_1+f_1)(w_2+f_2) + (w_2+f_2)(w_1+f_1) - t(f_1(w_2) + f_2(w_1))$$

which gives:

$$w_1w_2\ +\ w_1f_2\ +\ f_1w_2\ +\ f_1f_2\ +\ w_2w_1\ +\ w_2f_1\ +\ f_2w_1\ +\ f_2f_1\ -\ t(f_1(w_2)\ +\ f_2(w_1))$$

So we get elements like:

$$f_1w_2\ +\ w_2f_1\ -\ t(f_1(w_2))$$

Subtracting these type of elements from the elements of the generating form, we know that we have elements like:

$$wv + vw + fg + gf$$

and this of course gives us elements like

$$vw + wv\mbox{, and }fg + gf$$

A typical element of the Clifford algebra is of the form $x\ +\ \mathcal{I}$ where x is in T(V), the unit element is $1\ +\ \mathcal{I}$. The typical element of the ideal generated by $x\ +\ \mathcal{I}$, denoted $\langle x\ +\ \mathcal{I}\rangle$ is:

$$\sum _{i = 1} ^n (a_i\ +\ \mathcal{I})(x\ +\ \mathcal{I})(b_i\ +\ \mathcal{I})$$

So I want to find elements of the Clifford algebra $a_i\ +\ \mathcal{I}$, $b_i\ +\ \mathcal{I}$ such that:

$$\sum _{i = 1} ^n (a_i\ +\ \mathcal{I})(x\ +\ \mathcal{I})(b_i\ +\ \mathcal{I}) = 1\ +\ \mathcal{I}$$

I.e.:

$$\left [\left (\sum _{i = 1} ^n a_ixb_i\right )\ +\ \mathcal{I}\right ] = k\ +\ \mathcal{I}$$

for some non-zero k in $\mathbb{K}$. So we should be able to prove that for any x in T(V) not in the ideal, there are elements of T(V) a_i, b_i and a non-zero element of the field, k, satisfying:

$$\sum _{i = 1} ^n a_ixb_i - k \in \mathcal{I}$$

It's easy to see that if x is an element of the field, or of the form w, f, or w+f, this can be done. For example, if x is of the form w+f, choose some w' such that f(w') is non-zero, then let n=2, a₁ = b₂ = w', a₂ = b₁ = 1, k = tf(w') and we get:

w'(w + f)1 + 1(w + f)w' - tf(w')
= w'w + ww' + w'f + f'w - tf(w')

which is indeed an element of $\mathcal{I}$. x, however, can be much messier than any of these forms. Maybe this is a good start...?

 

[Hurkyl]

It might help to look at the structure of the Clifford algebra. Here's where I cheat and apply some of what little I know about them.

First off, I prefer yet simpler notation. I'm going to go whole hog and map the elements of W and W* all the way into Cl. So now, if I write w or f, I mean their image under the obvious map.

A Clifford algebra is generated by its degree-1 elements (the "vectors"). The algebraic relations between the vectors is given by a quadratic form Q on the underlying vector space.

That is, for any vector v, we have v² = Q(v).

Equivalently, it's characterized by its bilinear form. There's a symmetric product $v \cdot w = vw + wv$ (I've also seen it defined as half of that), and we have the relation $v \cdot w = \langle v, w \rangle$.

All algebraic relations between the vectors can be derived from the relation given by the quadratic form. Alternatively, they can be derived fom the one given by the bilinear form.

If you take an ordered basis for the underlying vector space, and view the Clifford algebra as a vector space, the corresponding basis for the Clifford algebra is:

All products of basis vectors (including the empty product, which is 1), such that:

(1) No basis vector is repeated. (It can be reduced via the quadratic form)
(2) The basis vectors appear in order. (If w comes after v, then wv can be reduced to vw via the bilinear form)

If i,j,k are basis vectors for V, then the basis for Cl(V) is simply:
1, i, j, k, ij, ik, jk, ijk

For example, ki = (ik + ki) - ik = <i, k> 1 + (-1) ik