Theorem 6.10 in Knapp's Basic Algebra: Exploring Bilinearity & Descending Maps

[Math Amateur]

I am reading Anthony W. Knapp's book: Basic Algebra in order to understand tensor products ... ...

I need some help with an aspect of Theorem 6.10 in Section 6 of Chapter VI: Multilinear Algebra ...

The text of Theorem 6.10 reads as follows:
View attachment 5405
View attachment 5406

The above proof mentions Figure 6.1 which is provided below ... as follows:View attachment 5407
In the above text, in the proof of Theorem 6.10 under "PROOF OF EXISTENCE" we read:

" ... ... The bilinearity of \(\displaystyle b\) shows that \(\displaystyle B_1\) maps \(\displaystyle V_0\) to \(\displaystyle 0\). By Proposition 2.25, \(\displaystyle B_1\) descends to a linear map \(\displaystyle B \ : \ V_1/V_0 \longrightarrow U\), and we have \(\displaystyle Bi = b\). "
My questions are as follows:


Question 1

Can someone please give a detailed demonstration of how the bilinearity of \(\displaystyle b\) shows that \(\displaystyle B_1\) maps \(\displaystyle V_0\) to \(\displaystyle 0\)?Question 2

Can someone please explain what is meant by "\(\displaystyle B_1\) descends to a linear map \(\displaystyle B \ : \ V_1/V_0 \longrightarrow U\)" and show why this is the case ... also showing why/how \(\displaystyle Bi = b\) ... ... ?

Hope someone can help ...

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

The above post mentions Proposition 2.25 so I am providing the text ... as follows:
View attachment 5408

============================================================*** EDIT 2 ***

After a little reflection it appears that the answer to my Question 2 above should "fall out" or result from matching the situation in Theorem 6.10 to that in Proposition 2.25 ... also I have noticed a remark of Knapp's following the statement of Proposition 2.25 which reads as follows:

View attachment 5409So that explains the language: "\(\displaystyle B_1\) descends to a linear map \(\displaystyle B \ : \ V_1/V_0 \longrightarrow U\)" ... ... BUT NOTE ...

I am having trouble applying Proposition 2.25 to Theorem 6.10 ... SO ... Question 2 remains a problem ... hope someone can help ...AND ... I remain perplexed over question 1 ...

Peter

 

[GJA]

Hi Peter,

Question 1

Can someone please give a detailed demonstration of how the bilinearity of \(\displaystyle b\) shows that \(\displaystyle B_1\) maps \(\displaystyle V_0\) to \(\displaystyle 0\)?

To show that $B_{1}$ maps $V_{0}$ to 0, it's enough to show that it sends basis elements to zero. Thus we need only compute $B_{1}$ on each of the four types of basis elements for $V_{0}$; these are the four expressions set apart in the text. For example, the first computation would be

$B_{1}((e_{1}+e_{2},f)-(e_{1},f)-(e_{2},f))=b(e_{1}+e_{2},f)-b(e_{1},f)-b(e_{2},f)=0,$

where the second equality follows from the bilinearity of $b$.

Question 2

Can someone please explain what is meant by "\(\displaystyle B_1\) descends to a linear map \(\displaystyle B \ : \ V_1/V_0 \longrightarrow U\)" and show why this is the case ... also showing why/how \(\displaystyle Bi = b\) ... ... ?

In Proposition 2.25 take $V=V_{1}$ and $W=U$, and the quotient space to be $V_{1}/V_{0}$; i.e. $U=V_{0}$ (I am trying to match the notation in Proposition 2.25, so please do not take the above to mean $W=V_{0}$). Then we have $Bq=B_{1}$. Now a computation will give you that $B\iota = b$; i.e. start with $B\iota (e,f)$, then see if you can use the result of Proposition 2.25 together with the definition of $\iota$, $q$, and $B_{1}$ to obtain $B\iota(e,f)=b(e,f).$

Let me know if this is still unclear. Good luck!

 

[Math Amateur]

Hi Peter,
To show that $B_{1}$ maps $V_{0}$ to 0, it's enough to show that it sends basis elements to zero. Thus we need only compute $B_{1}$ on each of the four types of basis elements for $V_{0}$; these are the four expressions set apart in the text. For example, the first computation would be

$B_{1}((e_{1}+e_{2},f)-(e_{1},f)-(e_{2},f))=b(e_{1}+e_{2},f)-b(e_{1},f)-b(e_{2},f)=0,$

where the second equality follows from the bilinearity of $b$.
In Proposition 2.25 take $V=V_{1}$ and $W=U$, and the quotient space to be $V_{1}/V_{0}$; i.e. $U=V_{0}$ (I am trying to match the notation in Proposition 2.25, so please do not take the above to mean $W=V_{0}$). Then we have $Bq=B_{1}$. Now a computation will give you that $B\iota = b$; i.e. start with $B\iota (e,f)$, then see if you can use the result of Proposition 2.25 together with the definition of $\iota$, $q$, and $B_{1}$ to obtain $B\iota(e,f)=b(e,f).$

Let me know if this is still unclear. Good luck!

Thanks GJA ... most helpful ...

BUT ... I am still a bit perplexed with showing why/how \(\displaystyle Bi = b\) ...I will try to explain my issues ...

Now Proposition 2.25 has a situation as shown below in Figure 1 ...View attachment 5414At this point you suggest putting \(\displaystyle V = V_1\) and \(\displaystyle W = U\) giving us the following situation ...
View attachment 5415Now ... for us to get \(\displaystyle Bq = B_1\) we require Figure 2 to look like Figure 3 below ... not quite sure how the assignment of maps in this way is justified, mind ...
...

https://www.physicsforums.com/attachments/5416
I am bothered by what is the exact nature of U ... (what is the form of the elements of U ... and why are they a particular form) ... .. and I am also worried and unsure about justifying that \(\displaystyle B_1\) is a map between \(\displaystyle V_1\) and \(\displaystyle U\) ... ...Can you elaborate and clarify ... ...


To further indicate my worries ... consider the following ...Knapp indicates that \(\displaystyle U\) is the codomain of a map \(\displaystyle b\) given by:\(\displaystyle b \ : \ E \times F \longrightarrow U\) ... ... as in Figure 6.1 shown below:https://www.physicsforums.com/attachments/5417
But Knapp also defines \(\displaystyle U\) as the codomain of the map \(\displaystyle B_1\) which Knapp defines as follows:\(\displaystyle B_1 ( \sum_{ (finite) } c_i ( e_i , f_i ) ) = \sum_{ (finite) } c_i b ( e_i , f_i )\)
Now, for the above to 'work', the codomain of \(\displaystyle b\) must be the same as the codomain of \(\displaystyle B_1\) ... ... BUT ... ... is this the case? ... ... and if it is how do we show this rigorously ...

Note also that when you write:

" ... ... Then we have $Bq=B_{1}$. Now a computation will give you that $B\iota = b$; i.e. start with $B\iota (e,f)$, then see if you can use the result of Proposition 2.25 together with the definition of $\iota$, $q$, and $B_{1}$ to obtain $B\iota(e,f)=b(e,f).$ ... ... "I am unsure of exactly how we get from \(\displaystyle Bq = b\) to \(\displaystyle B \iota = b\) ... can you be more explicit ... ...


Hope you can help clarify the above issues ...

Peter

 

[GJA]

Hi Peter,

You're trying to prove a so-called "universal" property of the tensor product of two vector spaces (see Universal Property section of https://en.wikipedia.org/wiki/Tensor_product). What you're trying to prove is a semi-remarkable property of the tensor product - namely that there is a bilinear map from $E\times F$ to the tensor product such that if $U$ is any given vector space and $b$ is any bilinear map from $E\times F$ to $U$, then there is a unique linear map ($B$ in this author's notation) so that the diagram in Figure 6.1 commutes (i.e. $B\iota = b$). So $U$ is a given vector space and $b$ is a given bilinear map.

I am bothered by what is the exact nature of U ... (what is the form of the elements of U ... and why are they a particular form) ... .. and I am also worried and unsure about justifying that \(\displaystyle B_1\) is a map between \(\displaystyle V_1\) and \(\displaystyle U\) ... ...

As mentioned above, $U$ is a given vector space, nothing more is known about it. Since $b$ is a map whose codomain is $U$ and $B_{1}$ is defined via $b$, so too is $B_{1}$'s codomain $U$.

BUT ... I am still a bit perplexed with showing why/how \(\displaystyle Bi = b\) ...

To get you going on the computation, start with $B\iota(e,f) = B((e,f) + V_{0})$ where we have applied the definition of $\iota.$ There are a few more steps after this initial one, but not too many. Remember that using the definition of $q$ and $B_{1}$ will be needed as well as the result from Proposition 2.25.

Hopefully this helps shed some light on the matter.

 

[Math Amateur]

Hi Peter,

You're trying to prove a so-called "universal" property of the tensor product of two vector spaces (see Universal Property section of https://en.wikipedia.org/wiki/Tensor_product). What you're trying to prove is a semi-remarkable property of the tensor product - namely that there is a bilinear map from $E\times F$ to the tensor product such that if $U$ is any given vector space and $b$ is any bilinear map from $E\times F$ to $U$, then there is a unique linear map ($B$ in this author's notation) so that the diagram in Figure 6.1 commutes (i.e. $B\iota = b$). So $U$ is a given vector space and $b$ is a given bilinear map.
As mentioned above, $U$ is a given vector space, nothing more is known about it. Since $b$ is a map whose codomain is $U$ and $B_{1}$ is defined via $b$, so too is $B_{1}$'s codomain $U$.
To get you going on the computation, start with $B\iota(e,f) = B((e,f) + V_{0})$ where we have applied the definition of $\iota.$ There are a few more steps after this initial one, but not too many. Remember that using the definition of $q$ and $B_{1}$ will be needed as well as the result from Proposition 2.25.

Hopefully this helps shed some light on the matter.

Thanks yet again for help GJA ... but I need further help ...

Indeed, you write:

" ... ... To get you going on the computation, start with $B\iota(e,f) = B((e,f) + V_{0})$ where we have applied the definition of $\iota.$ There are a few more steps after this initial one, but not too many. Remember that using the definition of $q$ and $B_{1}$ will be needed as well as the result from Proposition 2.25. ... ... "


Sorry GJA ... cannot proceed anywhere from $B\iota(e,f) = B((e,f) + V_{0})$ ... unsure how definitions of \(\displaystyle q\) and \(\displaystyle B_1\) can be used to show \(\displaystyle B\iota(e,f) = b(e,f)\) or in other words that \(\displaystyle B\iota = b\) ... ...
But just to indicate my thinking ... here is the set of maps involved with \(\displaystyle B\) showing as existing ... because ... given the existence of the maps \(\displaystyle B_1\) and \(\displaystyle q\) ... Proposition 2.2.5 tells us \(\displaystyle B\) exists ... so fitting \(\displaystyle B\) into a diagram showing the other maps defined by Knapp in Theorem 10.6 we have the following diagram ... (is it correct?)
View attachment 5418
As mentioned above the linear map \(\displaystyle B\) exists as indicated in Figure 2:
View attachment 5419
So from Figure 3 below, given that \(\displaystyle B\) exists we can deduce that \(\displaystyle Bi = b\) ... as long as we can assume the commutativity of the diagram ... ... ( can we? why exactly?) ... ...
View attachment 5420

Now I know that arguing from diagrams is not a proof ... but it is an indication of how a proof might work ... is my 'diagrammatic argument' correct in as far as it goes ... ?


I would however be grateful if you could help me progress your argument ... that starts with $B\iota(e,f) = B((e,f) + V_{0})$ ... ...

Peter

 

[GJA]

Hi Peter,

The picture you drew in Figure 1 is correct and will get you where you want to go. You want to recognize that $(e,f)\in E\times F$ and $(e,f)\in V_{1}$. Then,

$B\iota(e,f)=B((e,f)+V_{0})=Bq(e,f)=B_{1}(e,f)=b(e,f)$

Since $(e,f)$ was arbitrary, $B\iota =b$. Hope this helps!