Rotations, Complex Matrices and Real Matrices - Proof of Tapp, Proposition 2.2

[Math Amateur]

I am reading Kristopher Tapp's book: Matrix Groups for Undergraduates.

I am currently focussed on and studying Section 1 in Chapter2, namely:

"1. Complex Matrices as Real Matrices".I need help in fully understanding the proof of Tapp's Proposition 2.2.

Proposition 2.2 and its proof read as follows:
https://www.physicsforums.com/attachments/3994
https://www.physicsforums.com/attachments/3995

In the above proof we read:

" ... ... The composition of the two down-arrows on the right is

\(\displaystyle R_{ \rho_n (B) } \circ R_{ \rho_n (A) } = R_{ \rho_n (A) \cdot \rho_n (B) } .
\)

On the other hand, since on the left

\(\displaystyle R_B \circ R_A = R_{AB}\),

this composition also equals \(\displaystyle R_{ \rho_n } (AB)\) ... ... ..."
My question is as follows:Why exactly/rigorously does the composition on the right equal \(\displaystyle R_{ \rho_n } (AB)\) ... ... ?
Hope someone can help ... ...

Peter

***EDIT*** I now have a a further question related to the following remark after the proof:
" ... ... It is easy to see that

\(\displaystyle \rho_n \ : \ M_n ( \mathbb{C} ) \rightarrow M_{2n} ( \mathbb{R} )\)

is injective but not surjective ... ... "
Can someone please explain why this is the case?Peter

 

[Fallen Angel]

Hi Peter,

The equality holds because the diagram is conmutative, so starting in one point and going for the path you choose the resultant application is the same.

For the second question, try to reflect a little bit on it, take $\rho_{1}$, and look why it is injective but not surjective, the general case is exactly the same, specially seing the block description Euge did in the other post for $\rho_{n}$

 

[Math Amateur]

Hi Peter,

The equality holds because the diagram is conmutative, so starting in one point and going for the path you choose the resultant application is the same.

For the second question, try to reflect a little bit on it, take $\rho_{1}$, and look why it is injective but not surjective, the general case is exactly the same, specially seing the block description Euge did in the other post for $\rho_{n}$

Hi Fallen Angel,

You write:

"The equality holds because the diagram is commutative, so starting in one point and going for the path you choose the resultant application is the same."

I am sorry Fallen Angel ... I am possibly being a bit slow ... but I do not see how the commutativity of the diagram leads to

\(\displaystyle R_{ \rho_n (A) \cdot \rho_n (B) } = R_{ \rho_n } (AB) \)
However ... ...

... ... I can see that going down the left of the diagram and then along to the bottom \(\displaystyle \mathbb{R}^{2n}\) gives\(\displaystyle R_A \circ R_B \circ f_n = R_{AB} \circ f_n\)and going to the same bottom \(\displaystyle \mathbb{R}^{2n}\) by going along the top and then down the right gives\(\displaystyle f_n \circ R_{ \rho_n (A)} \circ R_{ \rho_n (B)} = f_n \circ R_{ \rho_n (AB)}\)and so we have \(\displaystyle R_A \circ R_B \circ f_n = R_{AB} \circ f_n = f_n \circ R_{ \rho_n (A)} \circ R_{ \rho_n (B)} f_n \circ R_{ \rho_n (AB)} \)so:\(\displaystyle R_{AB} \circ f_n = f_n \circ R_{ \rho_n (AB)}\)But where do we go from here? (Indeed, maybe my whole approach to using the commutativity of the diagram is wrong-headed as it does not seem to be getting anywhere... ...)Can you help?Peter

 

[Fallen Angel]

You are near Peter, now use that $f_{n}$ is the canonicall isomorphism to conclude.

 

[Math Amateur]

You are near Peter, now use that $f_{n}$ is the canonicall isomorphism to conclude.

Sorry, Fallen Angel, I do not follow ... can you be more explicit and help further ...

Peter

 

[Fallen Angel]

Hi Peter,

This applications, $R_{AB}$ and $R_{\rho_{n}(AB)}$ are not between the same spaces, so when he says they are equal is omitting under the canonicall isomorphism, for example, when $n=1$, $f_{1}((R_{AB}(a+bi))=R_{\rho_{1}(AB)}(a,b)$.

----------------------------------------------------------------------------------------Re-reading the text I think we are misunderstanding what Tapp says, what he says is that $R_{\rho_{n}(A)\cdot \rho_{n}(B)}$ should be equal to $R_{\rho_{n}(AB)}$ not to $R_{AB}$ the key is that on the left you can skip the middle step by using $R_{AB}$ in order to go from top left to bottom left, hence you should be able to skip the middle step on the right by using the $\rho_{n}$ of the matrix you use for skip it in the left, in conclusion, is the same going in two steps than going in one, so

$R_{\rho_{n}(AB)}=R_{\rho_{n}(A)\cdot \rho_{n}(B)}$.Sorry for the mistakes above.

 

[Math Amateur]

Hi Peter,

This applications, $R_{AB}$ and $R_{\rho_{n}(AB)}$ are not between the same spaces, so when he says they are equal is omitting under the canonicall isomorphism, for example, when $n=1$, $f_{1}((R_{AB}(a+bi))=R_{\rho_{1}(AB)}(a,b)$.

----------------------------------------------------------------------------------------Re-reading the text I think we are misunderstanding what Tapp says, what he says is that $R_{\rho_{n}(A)\cdot \rho_{n}(B)}$ should be equal to $R_{\rho_{n}(AB)}$ not to $R_{AB}$ the key is that on the left you can skip the middle step by using $R_{AB}$ in order to go from top left to bottom left, hence you should be able to skip the middle step on the right by using the $\rho_{n}$ of the matrix you use for skip it in the left, in conclusion, is the same going in two steps than going in one, so

$R_{\rho_{n}(AB)}=R_{\rho_{n}(A)\cdot \rho_{n}(B)}$.Sorry for the mistakes above.

I am sorry, Fallen Angel I am not following the logic ... or at least missing points of it ...

My real trouble is in seeing the logic for Tapp's text when he says:

" ... since on the left \(\displaystyle R_B \circ R_A = R_{AB}\), this composition on the right also equals \(\displaystyle R_{ \rho_n (AB) }\) ... "

I do not follow this ... it seems very informal and unconvincing to me ... I certainly do not follow exactly why this should be true ... do you have a formal/rigorous argument as to why this is true ... ...

It would also help enormously if you could provide a formal/rigorous argument of

\(\displaystyle R_{ \rho_n (A) \cdot \rho_n (B) } = R_{ \rho_n } (AB) \)

It would be great if you are able to help in this ...

Peter

 

[Fallen Angel]

Hi Peter,

" ... since on the left \(\displaystyle R_B \circ R_A = R_{AB}\), this composition on the right also equals \(\displaystyle R_{ \rho_n (AB) }\) ... "

At first, I understood (and I think you too) that this means $R_{AB}"="R_{\rho_{n}(AB)}$, because they are "equal in some sense", that is, is the same application but seeing it as complex or real via the canonicall isomorphism $f_{n}$.

But in my last post I'm telling you that I think this is not the purpose of this sentence, the "composition on the right" is referring to $R_{\rho_{n}(A)\cdot \rho_{n}(B)}$ in the equation that is just above the sentence.

Then the following two questions are the same

" ... since on the left \(\displaystyle R_B \circ R_A = R_{AB}\), this composition on the right also equals \(\displaystyle R_{ \rho_n (AB) }\) ... "

I do not follow this

It would also help enormously if you could provide a formal/rigorous argument of

\(\displaystyle R_{ \rho_n (A) \cdot \rho_n (B) } = R_{ \rho_n } (AB) \)

So far, I also think you don't consider rigorous an argument that is written in words, when it's just a way to avoid writing too much, I think you expect something like the following

It's clear that

$R_{B}\circ R_{A}=R_{AB}$, right?

From here it should be also clear that

$R_{\rho_{n}(B)}\circ R_{\rho_{n}(A)}=R_{\rho_{n}(A)\cdot \rho_{n}(B)}$.

And we also know that $f_{n}\circ R_{B} \circ R_{A}= R_{\rho_{n}(B)} \circ R_{\rho_{n}(A)}\circ f_{n}$.

Using the above equalities

$f_{n}\circ R_{AB}= R_{\rho_{n}(A)\cdot \rho_{n}(B)}\circ f_{n}$

And now applying $\rho_{n}$ to $AB$ we reach

$f_{n}\circ R_{AB}= R_{\rho_{n}(AB)}\circ f_{n}$

And we got the equality we want.

Let me know if you still have problems.

 

[Math Amateur]

Hi Peter,
At first, I understood (and I think you too) that this means $R_{AB}"="R_{\rho_{n}(AB)}$, because they are "equal in some sense", that is, is the same application but seeing it as complex or real via the canonicall isomorphism $f_{n}$.

But in my last post I'm telling you that I think this is not the purpose of this sentence, the "composition on the right" is referring to $R_{\rho_{n}(A)\cdot \rho_{n}(B)}$ in the equation that is just above the sentence.

Then the following two questions are the sameSo far, I also think you don't consider rigorous an argument that is written in words, when it's just a way to avoid writing too much, I think you expect something like the following

It's clear that

$R_{B}\circ R_{A}=R_{AB}$, right?

From here it should be also clear that

$R_{\rho_{n}(B)}\circ R_{\rho_{n}(A)}=R_{\rho_{n}(A)\cdot \rho_{n}(B)}$.

And we also know that $f_{n}\circ R_{B} \circ R_{A}= R_{\rho_{n}(B)} \circ R_{\rho_{n}(A)}\circ f_{n}$.

Using the above equalities

$f_{n}\circ R_{AB}= R_{\rho_{n}(A)\cdot \rho_{n}(B)}\circ f_{n}$

And now applying $\rho_{n}$ to $AB$ we reach

$f_{n}\circ R_{AB}= R_{\rho_{n}(AB)}\circ f_{n}$

And we got the equality we want.

Let me know if you still have problems.

Hi Fallen Angel,

Thanks for your further help ...

I follow all you say until:" ... ... And now applying $\rho_{n}$ to $AB$ we reach

$f_{n}\circ R_{AB}= R_{\rho_{n}(AB)}\circ f_{n}$

... ... "Can you explain how applying \(\displaystyle \rho\) to \(\displaystyle AB\) gives

$f_{n}\circ R_{AB}= R_{\rho_{n}(AB)}\circ f_{n}$ ... ... ?

Hope you can help further in this ...

Peter
***EDIT***

I think I may have seen what you are driving at ... ...

If we consider the top square of the commutative diagram, then moving along both paths from the top left \(\displaystyle \mathbb{C}^n\) to the middle \(\displaystyle \mathbb{R}^{2n}\), we have:\(\displaystyle R_{ \rho_n (A) } \circ f_n = f_n \circ R_A\) ... ... ... (1)Now, if (1) is true for \(\displaystyle A \in M_n( \mathbb{C} )\) then it is true for \(\displaystyle AB\) where \(\displaystyle B \in M_n( \mathbb{C} )\) ... ... but is this true ... how do we know it must be true for \(\displaystyle AB\) ... not sure ...

... ... anyway ... go ahead ...

then

\(\displaystyle R_{ \rho_n (AB) } \circ f_n = f_n \circ R_{AB}\)

which, as you indicated, is what is required ...

Is that correct ...

Peter

 

[Fallen Angel]

By definition of $\rho_{n}$. It's the application that makes conmutative this diagram:

$\begin{array}{ccccc}& \Bbb{C}^{n} & \overset{f_{n}}{\longrightarrow} & \Bbb{R}^{2n} & \\
R_{AB}& \downarrow & & \downarrow & R_{\rho_{n}(AB)} \\
& \Bbb{C}^{n} & \overset{f_{n}}{\longrightarrow} & \Bbb{R}^{2n} &

\end{array}$

 

[Math Amateur]

By definition of $\rho_{n}$. It's the application that makes conmutative this diagram:

$\begin{array}{ccccc}& \Bbb{C}^{n} & \overset{f_{n}}{\longrightarrow} & \Bbb{R}^{2n} & \\
R_{AB}& \downarrow & & \downarrow & R_{\rho_{n}(AB)} \\
& \Bbb{C}^{n} & \overset{f_{n}}{\longrightarrow} & \Bbb{R}^{2n} &

\end{array}$

Yes, I just realized that a moment ago ...

Thank you so much for all your patient help ... much appreciated!

Peter