A Question about ##FG## modules

[MathLearner123]

Definition. Let $F$ be a field and $G$ be a group. An $FG$-module is a finite-dimensional vector space $V$ on which $G$ acts (from the right: $V \times G \to V, (v,g)\mapsto v\cdot g$) such that the next conditions hold:

1) $(v \cdot g)\cdot h = v \cdot (g \cdot h)$
2) $v \cdot e = v$
3) $(\lambda v) \cdot g = \lambda(v \cdot g)$
4) $(v+w)\cdot g = v \cdot g + w \cdot g$

for all $v, w \in V, \lambda \in F$ and $g,h \in G$.

And I have the next proposition which I don't understand:
Proposition. If $f:G \to GL_n(F)$ is a representation of the group $G$ and $V$ is a $n$-dimensional vector space over $F$, then $V$ is an $FG$-module with $v \cdot g = v \ f(g),$ for all $v \in V, g \in G$.

Ok, for $V = F^n$ I can think that $V$ is a line vector and I can multiply that line vector with the matrix $f(g)$ and obtain another line vector. But if $V$ is an arbitrary vector space over $F$, how the product $v \ f(g)$ makes sense? I thought that could be something in coordinates but I still can't figure it out...

 

[martinbn]

Yes, pick a basis of $V$ and identify it with $F^n$.

 

[fresh_42]

The words representation, operation, and module all describe the same thing from different perspectives. The change from $G$ to $FG$ only reflects that we want a distributive structure on the scalars, not just a multiplication so that it better matches the vector space structure on the representation space $V$. It extends the scalars in $G$ to elements in $FG$ which can be added and multiplied like ordinary scalars in a field could.

 

[mathwonk]

I think they meant GL(V). try not to get hung up by trivial stuff like this. everyone makes mistakes.

 

[MathLearner123]

Thanks for all the answers. I understand that I need to put $GL(V)$. But I think that it's sufficient to add to the statement of the proposition that $\{v_1, \ldots, v_n\}$ is a fixed basis for $V$ and then I can take $v \cdot g$ to be the unique vector from $v$ which have the coordinates given by product of coordinate vector of $v$ (because I will have an isomorphism between $V$ and $F^n$) and the matrix $f(g)$. I think the statement would be ok, because each fixed basis for $V$ would give different $FG$-module structure on $V$. It's ok? What do you think?

 

[fresh_42]

Thanks for all the answers. I understand that I need to put $GL(V)$. But I think that it's sufficient to add to the statement of the proposition that $\{v_1, \ldots, v_n\}$ is a fixed basis for $V$ and then I can take $v \cdot g$ to be the unique vector from $v$ which have the coordinates given by product of coordinate vector of $v$ (because I will have an isomorphism between $V$ and $F^n$) and the matrix $f(g)$. I think the statement would be ok, because each fixed basis for $V$ would give different $FG$-module structure on $V$. It's ok? What do you think?

I think that $\operatorname{GL}_n(F)=\operatorname{GL}(F^n)$ are only two different notations of the same linear group. We also have $\operatorname{GL}_n(F)=\operatorname{GL}(V)$ if $V$ is an n-dimensional $F$-vector space.

Changing the basis does not change the module structure, i.e. is not different. Say we have $v=\alpha_1 b_1+\ldots + \alpha_n b_n$ in one basis $\{b_1,\ldots,b_n\}$ and $b_k=\sum_{p=1}^n c_{pk}B_p$ in a basis $\{B_1,\ldots,B_n\}.$ Then
$$
v\cdot g=\sum_{q=1}^n\alpha_q (b_q\cdot g)=\sum_{q=1}^n\alpha_q\left(\sum_{p=1}^n c_{pq}B_p\right)\cdot g=\sum_{q=1}^n\alpha_q\sum_{p=1}^n c_{pq}\left(B_p\cdot g\right)
$$
You get different formulas for $v\cdot g$ depending on which basis you use to express the vector $v$ but the operation itself did not change. We still have the same group homomorphism $f.$ It is as it always is when changing a basis: the matrix changes, the linear transformation does not.

A side note: Most authors write $v\cdot g$ from the other side as a left module $g.v$ so don't be surprised if you see it written the other way around somewhere.

 

[martinbn]

A side note: Most authors write $v\cdot g$ from the other side as a left module $g.v$ so don't be surprised if you see it written the other way around somewhere.

Yes, but one is a left action, the other a right action.

 

[MathLearner123]

I think that $\operatorname{GL}_n(F)=\operatorname{GL}(F^n)$ are only two different notations of the same linear group. We also have $\operatorname{GL}_n(F)=\operatorname{GL}(V)$ if $V$ is an n-dimensional $F$-vector space.

Changing the basis does not change the module structure, i.e. is not different. Say we have $v=\alpha_1 b_1+\ldots + \alpha_n b_n$ in one basis $\{b_1,\ldots,b_n\}$ and $b_k=\sum_{p=1}^n c_{pk}B_p$ in a basis $\{B_1,\ldots,B_n\}.$ Then
$$
v\cdot g=\sum_{q=1}^n\alpha_q (b_q\cdot g)=\sum_{q=1}^n\alpha_q\left(\sum_{p=1}^n c_{pq}B_p\right)\cdot g=\sum_{q=1}^n\alpha_q\sum_{p=1}^n c_{pq}\left(B_p\cdot g\right)
$$
You get different formulas for $v\cdot g$ depending on which basis you use to express the vector $v$ but the operation itself did not change. We still have the same group homomorphism $f.$ It is as it always is when changing a basis: the matrix changes, the linear transformation does not.

A side note: Most authors write $v\cdot g$ from the other side as a left module $g.v$ so don't be surprised if you see it written the other way around somewhere.

Sorry if I answer too late after you replied. Ok. What I've referred to by saying "different structure" was the fact that if I choose fixed vector $v \in V$ and a basis $\{b_1, \ldots, b_n\}$ then I can get $v \cdot g = v_1 \in V$. If I choose another basis, let it be $\{B_1, \ldots, B_n\}$, I can get $v \cdot g = v_2$ with $v_1 \neq v_2$. So a basis needs to be fixed in the statement of proposition to avoid this type of problem. It is true or I am wrong? Thanks!

 

[fresh_42]

Sorry if I answer too late after you replied. Ok. What I've referred to by saying "different structure" was the fact that if I choose fixed vector $v \in V$ and a basis $\{b_1, \ldots, b_n\}$ then I can get $v \cdot g = v_1 \in V$. If I choose another basis, let it be $\{B_1, \ldots, B_n\}$, I can get $v \cdot g = v_2$ with $v_1 \neq v_2$. So a basis needs to be fixed in the statement of proposition to avoid this type of problem. It is true or I am wrong? Thanks!

You are wrong. You get different numbers but the same result. It is as it is with any change of basis. If you change the basis, then already $v$ is written in two ways. So what is $v$ in your equation? Is it $v=\sum a_kb_k$ or is it $v=\sum a_kB_k$?

Take the group $G=\operatorname{SL}_n(\mathbb{R})$ of matrices with determinant one, and the real vector space of $2\times 2$ matrices with trace zero, i.e. matrices of the form $v=\begin{pmatrix}a&b\\c&-a\end{pmatrix}.$ Then define the operation $v.g=gvg^{-1}.$ You can take
$$
b_1=\begin{pmatrix}0&1\\0&0\end{pmatrix}\, , \,b_2=\begin{pmatrix}0&0\\1&0\end{pmatrix}\, , \,b_3=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\, , \,
$$
and
$$
B_1=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\, , \,B_2=\begin{pmatrix}0&2\\1&0\end{pmatrix}\, , \,B_3=\begin{pmatrix}3&0\\0&-3\end{pmatrix}\, , \,
$$
If you write down the equations, then you get different numbers because of different bases, but the operation is still the same.

Or take a rotation of the Euclidean plane as group, and the plane as a vector space. Changing the bases of this vector space changes the numbers, but not the rotation. $v_1=v_2$ but expressed differently. As is $v.$
$$
v=1\cdot \begin{pmatrix}1\\0\end{pmatrix}+0\cdot \begin{pmatrix}0\\1\end{pmatrix}=2\cdot\begin{pmatrix}2\\1\end{pmatrix}+1\cdot\begin{pmatrix}-3\\-2\end{pmatrix}
$$
The vector $v$ is the same, but its coordinates are $(1,0)$ in the first case and $(2,1)$ in the second. I wouldn't call that different because it suggests that $v$ has changed, but we only changed the scales of the plane, the basis vectors, i.e. the way we write $v.$ This is the same with $v.g.$ It does not change what happens, only how we write it.

 

[MathLearner123]

You are wrong. You get different numbers but the same result. It is as it is with any change of basis. If you change the basis, then already $v$ is written in two ways. So what is $v$ in your equation? Is it $v=\sum a_kb_k$ or is it $v=\sum a_kB_k$?

Take the group $G=\operatorname{SL}_n(\mathbb{R})$ of matrices with determinant one, and the real vector space of $2\times 2$ matrices with trace zero, i.e. matrices of the form $v=\begin{pmatrix}a&b\\c&-a\end{pmatrix}.$ Then define the operation $v.g=gvg^{-1}.$ You can take
$$
b_1=\begin{pmatrix}0&1\\0&0\end{pmatrix}\, , \,b_2=\begin{pmatrix}0&0\\1&0\end{pmatrix}\, , \,b_3=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\, , \,
$$
and
$$
B_1=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\, , \,B_2=\begin{pmatrix}0&2\\1&0\end{pmatrix}\, , \,B_3=\begin{pmatrix}3&0\\0&-3\end{pmatrix}\, , \,
$$
If you write down the equations, then you get different numbers because of different bases, but the operation is still the same.

Or take a rotation of the Euclidean plane as group, and the plane as a vector space. Changing the bases of this vector space changes the numbers, but not the rotation. $v_1=v_2$ but expressed differently. As is $v.$
$$
v=1\cdot \begin{pmatrix}1\\0\end{pmatrix}+0\cdot \begin{pmatrix}0\\1\end{pmatrix}=2\cdot\begin{pmatrix}2\\1\end{pmatrix}+1\cdot\begin{pmatrix}-3\\-2\end{pmatrix}
$$
The vector $v$ is the same, but its coordinates are $(1,0)$ in the first case and $(2,1)$ in the second. I wouldn't call that different because it suggests that $v$ has changed, but we only changed the scales of the plane, the basis vectors, i.e. the way we write $v.$ This is the same with $v.g.$ It does not change what happens, only how we write it.

Suppose that I have a representation of a group $G$, and $g \in G$ with $f(g) = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$. Let $V = \mathcal{M}_{1,2}(\mathbb{R})$ (line matrices) and take the bases $B_1 = \{(1,0), (0,1)\}$ and $B_2 = \{(3,4), (2,5)\}$. Also, take $v = (7,11) \in V$. Then, we can observe that the coordinates of $v$ are $$[v]_{B_1}=(7,11)$$ and $$[v]_{B_2}=\left(\frac{13}{7}, \frac{5}{7}\right).$$ So, we have that in the basis $B_1$, $[v \cdot g]_{B_1} = \left[(7,11) \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\right]_{B_1} = (25, 68)$, so $v \cdot g = 25 (1,0) + 68 (0,1) = (25,68)$ and in the basis $B_2$ we have $[v \cdot g]_{B_2} = \left[(\frac{13}{7},\frac{5}{7}) \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\right]_{B_2} = (\frac{31}{7}, \frac{80}{7})$, so $v \cdot g = \frac{31}{7} (3,4) + \frac{80}{7} (2,5) = (\frac{253}{7}, \frac{524}{7})$. So If I change the basis, I get different vectors in the plane.

 

[fresh_42]

Suppose that I have a representation of a group $G$, and $g \in G$ with $f(g) = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$. Let $V = \mathcal{M}_{1,2}(\mathbb{R})$ (line matrices) and take the bases $B_1 = \{(1,0), (0,1)\}$ and $B_2 = \{(3,4), (2,5)\}$. Also, take $v = (7,11) \in V$. Then, we can observe that the coordinates of $v$ are $$[v]_{B_1}=(7,11)$$ and $$[v]_{B_2}=\left(\frac{13}{7}, \frac{5}{7}\right).$$ So, we have that in the basis $B_1$, $[v \cdot g]_{B_1} = \left[(7,11) \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\right]_{B_1} = (25, 68)$, so $v \cdot g = 25 (1,0) + 68 (0,1) = (25,68)$ and in the basis $B_2$ we have $[v \cdot g]_{B_2} = \left[(\frac{13}{7},\frac{5}{7}) \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\right]_{B_2} = (\frac{31}{7}, \frac{80}{7})$, so $v \cdot g = \frac{31}{7} (3,4) + \frac{80}{7} (2,5) = (\frac{253}{7}, \frac{524}{7})$. So If I change the basis, I get different vectors in the plane.

It is (very) late here and you ask the wrong person when it comes to a change of basis. I tend to confuse which is which and what is what. Let me see if I can get it straight.

The general mistake in your calculation was that you didn't adjust the operation. Your $f$ is already dependent on the basis. Hence you applied $f$ written in basis one to vectors written in basis two.

I don't like upper indices, so let me write basis one as
$$\mathcal{B}=\left\{b_1=\begin{pmatrix}1\\0\end{pmatrix}\, , \,b_2=\begin{pmatrix}0\\1\end{pmatrix}\right\}$$
and basis two as
$$\mathcal{C}=\left\{c_1=\begin{pmatrix}3\\4\end{pmatrix}\, , \,c_2=\begin{pmatrix}2\\5\end{pmatrix}\right\}.$$
Now what happens under the operation of $f=\begin{pmatrix}2&5\\1&3\end{pmatrix}_\mathcal{B}$? It transforms
$$
b_1\longrightarrow 2b_1+5b_2\quad\text{ and }\quad b_2\longrightarrow 1b_1+3b_2
$$
This means that the other basis vectors are transformed by
\begin{align*}
c_1&=3b_1+4b_2 \longrightarrow 6b_1+15b_2+4b_1+12b_2=10b_1+27b_2=-\dfrac{4}{7}c_1+\dfrac{41}{7}c_2\\[10pt]
c_2&=2b_1+5b_2\longrightarrow 4b_1+10b_2+5b_1+15b_2=9b_1+25b_2=-\dfrac{5}{7}c_1+\dfrac{39}{7}c_2
\end{align*}
The operation $f$ written in the basis $\mathcal{C}$ is therefore given by $f=\dfrac{1}{7}\begin{pmatrix}-4&41\\-5&39\end{pmatrix}_\mathcal{C}$ and
\begin{align*}
v_\mathcal{C} f(g)&=\dfrac{1}{7}(13,5)_C \dfrac{1}{7} \begin{pmatrix}-4&41\\-5&39\end{pmatrix}_\mathcal{C} = \dfrac{1}{49}(-77,728)_\mathcal{C}=\dfrac{1}{7}(-11,104)_\mathcal{C} \\[10pt]
&=-\dfrac{11}{7}\begin{pmatrix}3\\4 \end{pmatrix}_\mathcal{B}+\dfrac{104}{7}\begin{pmatrix}2\\5\end{pmatrix}_\mathcal{B}=(25,68)_\mathcal{B}
\end{align*}
Matrix multiplication always depends on the basis it is written in and you used matrix multiplication as operation.

 

[MathLearner123]

It is (very) late here and you ask the wrong person when it comes to a change of basis. I tend to confuse which is which and what is what. Let me see if I can get it straight.

The general mistake in your calculation was that you didn't adjust the operation. Your $f$ is already dependent on the basis. Hence you applied $f$ written in basis one to vectors written in basis two.

I don't like upper indices, so let me write basis one as
$$\mathcal{B}=\left\{b_1=\begin{pmatrix}1\\0\end{pmatrix}\, , \,b_2=\begin{pmatrix}0\\1\end{pmatrix}\right\}$$
and basis two as
$$\mathcal{C}=\left\{c_1=\begin{pmatrix}3\\4\end{pmatrix}\, , \,c_2=\begin{pmatrix}2\\5\end{pmatrix}\right\}.$$
Now what happens under the operation of $f=\begin{pmatrix}2&5\\1&3\end{pmatrix}_\mathcal{B}$? It transforms
$$
b_1\longrightarrow 2b_1+5b_2\quad\text{ and }\quad b_2\longrightarrow 1b_1+3b_2
$$
This means that the other basis vectors are transformed by
\begin{align*}
c_1&=3b_1+4b_2 \longrightarrow 6b_1+15b_2+4b_1+12b_2=10b_1+27b_2=-\dfrac{4}{7}c_1+\dfrac{41}{7}c_2\\[10pt]
c_2&=2b_1+5b_2\longrightarrow 4b_1+10b_2+5b_1+15b_2=9b_1+25b_2=-\dfrac{5}{7}c_1+\dfrac{39}{7}c_2
\end{align*}
The operation $f$ written in the basis $\mathcal{C}$ is therefore given by $f=\dfrac{1}{7}\begin{pmatrix}-4&41\\-5&39\end{pmatrix}_\mathcal{C}$ and
\begin{align*}
v_\mathcal{C} f(g)&=\dfrac{1}{7}(13,5)_C \dfrac{1}{7} \begin{pmatrix}-4&41\\-5&39\end{pmatrix}_\mathcal{C} = \dfrac{1}{49}(-77,728)_\mathcal{C}=\dfrac{1}{7}(-11,104)_\mathcal{C} \\[10pt]
&=-\dfrac{11}{7}\begin{pmatrix}3\\4 \end{pmatrix}_\mathcal{B}+\dfrac{104}{7}\begin{pmatrix}2\\5\end{pmatrix}_\mathcal{B}=(25,68)_\mathcal{B}
\end{align*}
Matrix multiplication always depends on the basis it is written in and you used matrix multiplication as operation.

Thanks for your help and sorry if I bothered you with my questions!

 

[fresh_42]

Thanks for your help and sorry if I bothered you with my questions!

No problem. It is in general a bit confusing. As soon as we write a vector as $v=(\alpha_1,\ldots,\alpha_n)$ we automatically use a basis: the coordinates are the coefficients at the basis vectors, the components. And rotating a rectangle by 90° or turning the table by 90° yields the same result. So did we rotate a geometric figure or the coordinate system?

 

[MathLearner123]

No problem. It is in general a bit confusing. As soon as we write a vector as $v=(\alpha_1,\ldots,\alpha_n)$ we automatically use a basis: the coordinates are the coefficients at the basis vectors, the components. And rotating a rectangle by 90° or turning the table by 90° yields the same result. So did we rotate a geometric figure or the coordinate system?

I think that we rotate the coordinate system.

 

[fresh_42]

I think that we rotate the coordinate system.

The point is that rotating a pen by +45° and rotating the table by -45° yields the same position of the pen on the table. It only differs in the frame of the room, but there is no outer coordinate system 'room' in mathematics. We can either rotate the pen or the coordinate system 'table' in the opposite direction. The result for the system (pen, table) is the same.

This is why coordinate transformations in linear algebra are written as conjugations.
\begin{align*}
&V\quad\stackrel{\varphi}{\longrightarrow } V\\
&\uparrow S\quad \;\;\;\;\downarrow S^{-1}\\
&V'\quad\stackrel{\psi}{\longrightarrow } V'
\end{align*}
If we want to write the linear transformation $\varphi$ of a vectors space $V$ in a different basis $V'$ then we have $\psi = S^{-1}\varphi S.$ We get different numbers for the matrices of $\varphi$ and $\psi$ but the transformation is the same. That is why we always need to know according to which basis a vector or a matrix is written. $(7,-1)$ means something different in basis $\left\{(2,0),0,2)\right\}$ and in basis $\left\{(3,4),(2,5)\right\}.$ And this is why it becomes confusing. We turn around in cycles since the way I wrote the basis vectors, e.g. $(3,4),$ isn't explained.

The solution to this paradox is:

Given a Cartesian coordinate system, perpendicular axis with a unit length $1$ on both of them. This allows us to define $b_1=(2,0), b_2=(0,2),c_1=(3,4),c_2=(2,5)$ unambiguously. The numbers of $v=(7,-1)$ are now not coordinates anymore, they are components relative to a basis.
$$
v=1\cdot b_1 -1\cdot b_2
$$
or
$$
v=1\cdot c_1 -1 \cdot c_2 .
$$
This means we get the two different arrows $v=(14,-2)$ or $v=(19,23)$ in our drawn Cartesian coordinate system although the components remained $(7,-1)$ in both cases. That is we actually dealt with three bases: $\left\{b_1,b_2\right\}\, ,\, \left\{c_1,c_2\right\}$ and what is drawn on the paper as $\left\{(1,0),(0,1)\right\}.$ The result are three different arrows although we never changed the notation $(7,-1).$ And the same holds the other way around. We can draw one arrow on the paper and a) measure its Cartesian coordinates on the paper, b) express it in the basis $\left\{b_1,b_2\right\},$ or c) express it in the basis $\left\{c_1,c_2\right\}.$ Then we have the same arrow but three different sets of coordinates depending on the basis we have chosen.

If we do not name or worse, do not know, which of these are meant, then it is confusing and chances are that two people have different perspectives of the same notation $(7,-1).$






The better notation would be