Proving Grassmann's Algebra Using Free Vector Spaces

[ilia1987]

I currently self study from the book "A Course in Modern Mathematical Physics" by Peter Szekeres, and I'm currently reading the chapter on tensors, which he defines using the concept of Free Vector Spaces.
He gives a re-definition of Grassmann's algebras introduced in the previous section by using the concept of free algebra. And I had the following problem while reading the text:
(This is a Citation:) "
Let $$\mathcal{F}(V)$$ be the free associative algebra over a real vector space $$V$$, and let $$\mathcal{S}$$ be the ideal generated by all elements of $$\mathcal{F}(V)$$ of the form $$u \otimes T \otimes v + v \otimes T \otimes u$$ where $$u,v\in V$$ and $$T\in \mathcal{F}(V)$$. The general element of $$\mathcal{S}$$ is $$S \otimes u \otimes T \otimes v \otimes U + S \otimes v \otimes T \otimes u \otimes U$$ where $$u,v\in V$$ and $$S,T,U\in \mathcal{F}(V)$$
"
My question is: Why can every element be expressed in this way?
An Ideal is first of all a Vector Subspace, right? So the sum of any two such "general elements" is supposed to be a general element too.
In other words, how do I prove that :
$$S_1 \otimes u_1 \otimes T_1 \otimes v_1 \otimes U_1 + S_1 \otimes v_1 \otimes T_1 \otimes u_1 \otimes U_1 \ \ \ \ + \ \ \ \ S_2 \otimes u_2 \otimes T_2 \otimes v_2 \otimes U_2 + S_2 \otimes v_2 \otimes T_2 \otimes u_2 \otimes U_2 \ \ \ \ = \ \ \ \ S_3 \otimes u_3 \otimes T_1 \otimes v_3 \otimes U_3 + S_3 \otimes v_3 \otimes T_3 \otimes u_3 \otimes U_3$$
?

Thank you for your time.

 

[HallsofIvy]

I currently self study from the book "A Course in Modern Mathematical Physics" by Peter Szekeres, and I'm currently reading the chapter on tensors, which he defines using the concept of Free Vector Spaces.
He gives a re-definition of Grassmann's algebras introduced in the previous section by using the concept of free algebra. And I had the following problem while reading the text:
(This is a Citation:) "
Let $$\mathcal{F}(V)$$ be the free associative algebra over a real vector space $$V$$, and let $$\mathcal{S}$$ be the ideal generated by all elements of $$\mathcal{F}(V)$$ of the form $$u \otimes T \otimes v + v \otimes T \otimes u$$ where $$u,v\in V$$ and $$T\in \mathcal{F}(V)$$. The general element of $$\mathcal{S}$$ is $$S \otimes u \otimes T \otimes v \otimes U + S \otimes v \otimes T \otimes u \otimes U$$ where $$u,v\in V$$ and $$S,T,U\in \mathcal{F}(V)$$
"
My question is: Why can every element be expressed in this way?

Because that is what "generated by" means!

An Ideal is first of all a Vector Subspace, right So the sum of any two such "general elements" is supposed to be a general element too.
In other words, how do I prove that :
$$S_1 \otimes u_1 \otimes T_1 \otimes v_1 \otimes U_1 + S_1 \otimes v_1 \otimes T_1 \otimes u_1 \otimes U_1 \ \ \ \ + \ \ \ \ S_2 \otimes u_2 \otimes T_2 \otimes v_2 \otimes U_2 + S_2 \otimes v_2 \otimes T_2 \otimes u_2 \otimes U_2 \ \ \ \ = \ \ \ \ S_3 \otimes u_3 \otimes T_1 \otimes v_3 \otimes U_3 + S_3 \otimes v_3 \otimes T_3 \otimes u_3 \otimes U_3$$
?

Thank you for your time.

 

[ilia1987]

I'm sorry, but if that's what generated means, the general element should be of the form:

$$\forall \ x\in \mathcal{S} , \ x=\displaystyle\sum_r (S_r \otimes v_r \otimes T_r \otimes u_r \otimes U_r \ + \ S_r \otimes u_r \otimes T_r \otimes v_r \otimes U_r)$$

where $$S_r,T_r,U_r \in \mathcal{F}(V)$$ and $$u_r,v_r\in V$$.
The question is whether any sum of this form can be expressed as a single element of the same form.

 

[Goldbeetle]

Are you still reading that book? I am, and I'm just a few lines before the part you were discussing in the first message of this thread. Maybe we can exchange some comments.

 

[blue_raver22]

It is interesting to see that you are self-studying from a book on modern mathematical physics. Grassmann's algebra is a fundamental concept in this field, so it is great that you are exploring it further.

To answer your question, let's first define what an ideal is in the context of free vector spaces. An ideal is a subspace that is closed under multiplication by elements of the free vector space. In other words, if we have an element in the ideal and we multiply it by any element in the free vector space, the result is still in the ideal.

Now, let's take a closer look at the elements in the ideal \mathcal{S}. We can see that they are of the form S \otimes u \otimes T \otimes v \otimes U, where S, T, U are elements of the free vector space \mathcal{F}(V) and u, v are elements of the vector space V. This means that if we multiply any element in \mathcal{S} by an element in \mathcal{F}(V), the resulting element will still be in \mathcal{S}. This is because the multiplication of elements in \mathcal{F}(V) follows the associative property, so the order of multiplication does not matter.

Now, let's go back to your question. You are asking why every element can be expressed in the form given in the citation. This is because the ideal \mathcal{S} is generated by these specific elements. This means that any element in \mathcal{S} can be written as a linear combination of these specific elements. And as we established earlier, these specific elements are closed under multiplication by elements in \mathcal{F}(V), so any linear combination of them will still be in \mathcal{S}.

To prove the equality that you have stated, we can use the fact that \mathcal{S} is an ideal. This means that if we have two elements in \mathcal{S}, their sum will also be in \mathcal{S}. In other words, the left-hand side of the equality can be written as a linear combination of the elements in \mathcal{S}. And since \mathcal{S} is generated by the specific elements mentioned in the citation, we can rewrite this linear combination using those elements. This will give us the right-hand side of the equality.

I hope this