Tensor Algebras - Dummit and Foote, Section 11.5

[Math Amateur]

I am reading Dummit and Foote: Abstract Algebra (Third Edition) ... and am focused on Section 11.5 Tensor Algebras. Symmetric and Exterior Algebras ...

In particular I am trying to understand Theorem 31 but at present I am very unsure about how to interpret the theorem and need some help in understanding the basic form of the elements involved and the mechanics of computations ... so would appreciate any help however simple ...

Theorem 31 and its proof read as follows:
View attachment 5555

My (rather simple) questions are as follows:Question 1

In the above text from D&F we read the following:" ... ... \(\displaystyle \mathcal{T} (M)\) is an \(\displaystyle R\)-Algebra containing \(\displaystyle M\) with multiplication defined by the mapping:

\(\displaystyle ( m_1 \otimes \ ... \ \otimes m_i ) ( m'_1 \otimes \ ... \ \otimes m'_j ) = m_1 \otimes \ ... \ \otimes m_i \otimes m'_1 \otimes \ ... \ \otimes m'_j \)

... ... ... "


... my questions are as follows:

What do the distributive laws look like in this case ... and would sums of elements be just formal sums ... or would we be able to add elements in the same sense as in the ring \(\displaystyle \mathbb{Z}\) where the sum \(\displaystyle 2+3\) gives an entirely different element \(\displaystyle 5\) ... ?

Further, how do we know that with respect to multiplication \(\displaystyle \mathcal{T}^{i} (M) \ \mathcal{T}^{j} (M) \subseteq \mathcal{T}^{i+j} (M)\) ... ... ?

Question 2

In the proof we read the following:"The map

\(\displaystyle \underbrace{ M \times M \times \ ... \ \times M }_{ i \ factors} \times \underbrace{ M \times M \times \ ... \ \times M }_{ j \ factors} \longrightarrow \mathcal{T}^{i+j} (M) \)

defined by

\(\displaystyle (m_1, \ ... \ , m_i, m'_1, \ ... \ , m'_j) \mapsto m_1 \otimes \ ... \ ... \ \otimes m_i \otimes m'_1 \otimes \ ... \ ... \ \otimes m'_j \)

is \(\displaystyle R\)-multilinear, so induces a bilinear map \(\displaystyle \mathcal{T}^{i} (M) \times \mathcal{T}^{j} (M)\) to \(\displaystyle \mathcal{T}^{i+j} (M)\) ... ... "My questions are:

... what does the multlinearity of the above map look like ... ?

and

... how do we demonstrate that the above map induces a bilinear map \(\displaystyle \mathcal{T}^{i} (M) \times \mathcal{T}^{j} (M)\) to \(\displaystyle \mathcal{T}^{i+j} (M)\) ... ... ? How/why is this the case... ?Hope someone can help ...

Peter
============================================================*** EDIT ***

To clarify my basic issue/problem with the Theorem ... it concerns the nature of elements of \(\displaystyle \mathcal{T} (M)\)Regarding this issue ... we have that ...\(\displaystyle \mathcal{T} (M) = R \oplus \mathcal{T}^1 (M) \oplus \mathcal{T}^2 (M) \oplus \mathcal{T}^1 (M) \oplus \ ... \ ... \ ... \)

which seems to suggest that an element of \(\displaystyle \mathcal{T} (M)\) is of the form

\(\displaystyle (r, m_1, m_2 \otimes m_3, m_4 \otimes m_5 \otimes m_6, \ ... \ ... \ ... \ ) \)

where only a finite number of terms are different from zero (finite support) ... ...BUT ... ... ... the definition of multiplication for \(\displaystyle \mathcal{T} (M)\) seems to imply that elements of \(\displaystyle \mathcal{T} (M)\) are of the form:\(\displaystyle m_1 \otimes m_2 \otimes \ ... \ ... \ \otimes m_i \)
?Can someone please clarify ...?Peter

 

[Deveno]

It's sort of like polynomials, the tensors of rank $k$ are like $x^k$ terms: when we ADD, they stay separate, when we MULTIPLY (in this case the "multiplication" is tensoring), they "jump ranks".

So, for example:

$[v_1 + (v_2\otimes v_3 + v_4 \otimes v_5)]\otimes [v_6\otimes v_7\otimes v_8]$

$= (v_1 \otimes v_6 \otimes v_7 \otimes v_8) + (v_1\otimes v_3 \otimes v_6 \otimes v_7 \otimes v_8 + v_4\otimes v_5\otimes v_6 \otimes v_7 \otimes v_8)$

 

[Math Amateur]

It's sort of like polynomials, the tensors of rank $k$ are like $x^k$ terms: when we ADD, they stay separate, when we MULTIPLY (in this case the "multiplication" is tensoring), they "jump ranks".

So, for example:

$[v_1 + (v_2\otimes v_3 + v_4 \otimes v_5)]\otimes [v_6\otimes v_7\otimes v_8]$

$= (v_1 \otimes v_6 \otimes v_7 \otimes v_8) + (v_1\otimes v_3 \otimes v_6 \otimes v_7 \otimes v_8 + v_4\otimes v_5\otimes v_6 \otimes v_7 \otimes v_8)$

Thanks for the help Deveno ... that clarified things somewhat ...

Are you able to shed light on my issue with the form and nature of elements of

... my statement regarding this issue is as follows: (repeating previous edit) ..

... ... To clarify my basic issue/problem with the Theorem ... it concerns the nature of elements of \(\displaystyle \mathcal{T} (M)\)Regarding this issue ... we have that ...\(\displaystyle \mathcal{T} (M) = R \oplus \mathcal{T}^1 (M) \oplus \mathcal{T}^2 (M) \oplus \mathcal{T}^1 (M) \oplus \ ... \ ... \ ... \)

which seems to suggest that an element of \(\displaystyle \mathcal{T} (M)\) is of the form

\(\displaystyle (r, m_1, m_2 \otimes m_3, m_4 \otimes m_5 \otimes m_6, \ ... \ ... \ ... \ ) \)

where only a finite number of terms are different from zero (finite support) ... ...BUT ... ... ... the definition of multiplication for \(\displaystyle \mathcal{T} (M)\) seems to imply that elements of \(\displaystyle \mathcal{T} (M)\) are of the form:\(\displaystyle m_1 \otimes m_2 \otimes \ ... \ ... \ \otimes m_i \) ... ... ?

Hope you can clarify ...

Peter

 

[Deveno]

Thanks for the help Deveno ... that clarified things somewhat ...

Are you able to shed light on my issue with the form and nature of elements of

... my statement regarding this issue is as follows: (repeating previous edit) ..

... ... To clarify my basic issue/problem with the Theorem ... it concerns the nature of elements of \(\displaystyle \mathcal{T} (M)\)Regarding this issue ... we have that ...\(\displaystyle \mathcal{T} (M) = R \oplus \mathcal{T}^1 (M) \oplus \mathcal{T}^2 (M) \oplus \mathcal{T}^1 (M) \oplus \ ... \ ... \ ... \)

which seems to suggest that an element of \(\displaystyle \mathcal{T} (M)\) is of the form

\(\displaystyle (r, m_1, m_2 \otimes m_3, m_4 \otimes m_5 \otimes m_6, \ ... \ ... \ ... \ ) \)

where only a finite number of terms are different from zero (finite support) ... ...BUT ... ... ... the definition of multiplication for \(\displaystyle \mathcal{T} (M)\) seems to imply that elements of \(\displaystyle \mathcal{T} (M)\) are of the form:\(\displaystyle m_1 \otimes m_2 \otimes \ ... \ ... \ \otimes m_i \) ... ... ?

Hope you can clarify ...

Peter

Well, both are right, and both are wrong.

We work in the direct sum of ALL ranks of tensors, so we can express sums of tensors of differing rank. These are "formal" sums, and as such we could just regard them as (finite) sequences of tensors indexed by rank. The analogy with polynomials is quite apt, we can regard polynomials as sequences of their coefficients.

But we can multiply tensors (using...the tensor product), and this multiplication takes place just as in the second expression. To get a "full" product, we have to "distribute" such as sum in two ways:

Once (within a rank) over the sums of "simple tensors" (tensors of a given rank that involve no formal sums, they just have "one term", like $v_1 \otimes v_2 \otimes v_3$ for a 3-tensor), and once (over all ranks) by collecting together all the sums of tensors of a given rank "in its proper slot" in the direct sum.

It would be "ugly" to write out a "fully general" element of $\mathcal{T}(M)$ of degree $k$, unless $k$ was quite small, say 2 or 3. The closest analogy I can think of would be polynomials in a (countably) infinite set of variables. For example:

$x^3,x^2z, xyz$ are all homogeneous polynomials of degree 3, and there's really no simplification of a linear combination of them. Similarly, we have a rather large array of homogeneous polynomials of degree 2:

$x_i^2, x_ix_j$ for any $i,j \in \Bbb N$.

So a "fully general" polynomial of say, degree 4, would be possibly tremendously long.