# Group Action - Permutation on the Polynomial

I'm trying to check the permutation on the polynomial is a Group Action, but I'm not getting the second axiom. I'm following my lecturer's work --- Examples 2.1 and 2.6 on page 5 on http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/gpaction.pdf --- I post this first. Can someone please spot the mistake? Thanks. Lecturer did: For $p \in S_n$ and $\textbf{v} = (c_1,c_2,\cdots,c_n) \in \mathbb{R^n}$, define $p \cdot \textbf{v} := (c_{p(1)},,\cdots,c_{p(n)})$. Check $$p_2 \cdot (\color{green}{p_1 \cdot (v)}) \overset{?}{\mathop{=}}\ (p_2 \cdot p_1)(v) \tag{\spadesuit}$$

LHS = $\color{maroon}{p_2} \cdot \color{green}{(c_{p_1(1)},,\cdots,c_{p_1(n)})} = \color{green}{{(c_{p_1(\color{maroon}{{p_2}(1)})}}},\cdots,\color{green}{{c_{p_1(\color{maroon}{{p_2}(n)})})} = (c_{(p_1{{p_2})(1)}},\cdots,c_{(p_1{{p_2})(n)}})}$ $\text{since$S_n$is a group so has associativity.}$

$= (p_1 \cdot p_2)(v) \neq RHS$. Hence above is NOT a group action.

I tried: Define $p \cdot f(x_1, \cdots,x_n) := f(x_{p(1)},,\cdots,x_{p(n)})$. Check this is a group action.

LHS of $(\spadesuit) = \color{maroon}{p_2} \cdot \color{green}{(x_{p_1(1)},,\cdots,c_{x_1(n)})} = \color{green}{{f(x_{p_1(\color{maroon}{{p_2}(1)})}}},\cdots,\color{green}{{x_{p_1(\color{maroon}{{p_2}(n)})})} = f(x_{(p_1{{p_2})(1)}},\cdots,x_{(p_1{{p_2})(n)}})} = (p_1 \cdot p_2)(v) \neq RHS$

Hence the above is NOT a group action?

This is tricky - the two cases look the same, but they're not. The first one is a right action $v \cdot (p_1 \cdot p_2) = (v \cdot p_1) \cdot p_2$, while the second one is a left action $p_1 \cdot (p_2 \cdot f) = (p_1 \cdot p_2) \cdot f$. To see why, consider these 2 permutations:

$$p_1(1) = 1, p_1(2) = 3, p_1(3) = 2$$ and $$p_2(1) = 3, p_2(2) = 2, p_2(3) = 1.$$

Let's write out explicitly what the actions are in the two cases to see the difference.

First, let's work out what the compositions $p_1 \cdot p_2$ and $p_2 \cdot p_1$ are.

The composition $p_1 \cdot p_2$ is:

$$p_1(p_2(1)) = 2, p_1(p_2(2)) = 3, p_1(p_2(3)) = 1$$

while the composition $p_2 \cdot p_1$ is:

$$p_2(p_1(1)) = 3, p_2(p_1(2)) = 1, p_2(p_1(3)) = 2.$$

Observe that they are not the same. We will use these later.

Now let's look at the 2 actions. The first action is on vectors. By the definition of the first action, $$p_1 \cdot (v_1, v_2, v_3) = (v_{p_1(1)}, v_{p_1(2)}, v_{p_1(3)}) = (v_1, v_3, v_2).$$ In words: $p_1$ acting on a vector interchanges the second and third coordinates. Similarly, $$p_2 \cdot (v_1, v_2, v_3) = (v_3, v_2, v_1)$$ In words: $p_2$ acting on a vector interchanges the first and third coordinates.

(In this situation, I find that thinking in words reduces the confusion: $p_1$ interchanges the second and third coordinates, not $v_2$ and $v_3$. You'll see the difference below.)

Thus, $$p_1 \cdot (p_2 \cdot v) = p_1 \cdot (p_2 \cdot (v_1, v_2, v_3)) = p_1 \cdot (v_3, v_2, v_1) = (v_3, v_1, v_2).$$ (If the last equality seems wrong, use the "words" description of $p_1$: Interchange the second and third coordinates.) Now the rightmost term above is $(p_2 \cdot p_1) \cdot v$, not $(p_1 \cdot p_2) \cdot v$. (Use the calculation of $p_2 \cdot p_1$ above.)

Conclusion: For the first action, on vectors, $p_1 \cdot (p_2 \cdot v) = (p_2 \cdot p_1) \cdot v$. So this is not a left action, but a right action.

Now look at the second action, which is not on vectors, but on real-valued functions on the set of all vectors. By definition, $$(p_1 \cdot f)(x_1, x_2, x_3) = f(x_{p_1(1)}, x_{p_1(2)}, x_{p_1(3)}) = f(x_1, x_3, x_2).$$

In words: To evaluate $p_1$ applied to a function at a vector, interchange the second and third coordinates of the vector, then apply the function.

Similarly, for $p_2$, the words version is: To evaluate $p_2$ applied to a function at a vector, interchange the first and third coordinates of the vector, then apply the function.

So what is $p_1 \cdot (p_2 \cdot f)$, applied to a vector? It is $$(p_1 \cdot (p_2 \cdot f))(x_1, x_2, x_3) = (p_2 \cdot f)(x_1, x_3, x_2) = f(x_2, x_3, x_1).$$

(Again, using the "words" description may reduce the confusion.) Now is this last term $p_1 \cdot p_2$ or $p_2 \cdot p_1$ applied to $f$? It is the former, as you can see from the calculation of $p_1 \cdot p_2$ above.

Conclusion: $p_1 \cdot (p_2 \cdot f) = (p_1 \cdot p_2) \cdot f$. This one is a left action.

I hope this clears up why these two cases aren't the same. Of course, I haven't yet proven that these are actions in general (I've only illustrated it for the particular $p_1$ and $p_2$ above), but hopefully this will give you the right idea for the general proof.

The key point is to remember that with these definitions, we are permuting the coordinates of the vector according to $p$, rather than the indices of the $v$'s according to $p$. If we did the latter instead, then the left and right action cases above would be reversed. See alias vs alibi for more discussion of this point.

• Thanks. I'll try this and get back to this post later. – Group Theory Dec 24 '12 at 23:31

I don't get it: if the action is

$$p\cdot v:=\left(c_{p(1)},...,c_{p(n)}\right)$$

then it must be

$$p_2\left(c_{p_1(1)},...,c_{p_1(n)}\right):=\left(c_{p_2(p(1))},...,c_{p_2(p_1(n))}\right)=\left(c_{p_2\circ p_1(1)},...,c_{p_2\circ p_1(n)}\right)$$

Why do you, and your lecturer, interchange the roles of the index between $\,p_1\,$ and $\,p_2\,$?

With the above I think you won't have problems...

• No, the roles have to be interchanged. The definition of the operation is $(p \cdot v)_i = v_{p(i)}$. Then $(p_2 \cdot (p_1 \cdot v))_i = (p_1 \cdot v)_{p_2(i)} = v_{p_1(p_2(i))} = ((p_1 \cdot p_2) \cdot v)_i$. As defined, this is a right action, not a left action. It would be more natural to write $v \cdot p$ for the action, and then we would have the natural looking law $(v \cdot p_1) \cdot p_2 = v \cdot (p_1 \cdot p_2)$. – Ted Dec 24 '12 at 22:51
• Thanks for posting back. I've uploaded a screenshot of the exact note. – Group Theory Dec 24 '12 at 23:30
• @Ted, I did not do $\,\left(p_2\cdot(p_1\cdot v)\right)_i\,$, but as it's clearly written there $\,p_2(c_{p(i)})\,$ , which is the other side. . Nevertheless I understand your point and see the inversing of the indexes of $\,p_i\,$ and, thus, to avoid problems (right-left action and etc.), to define, from the left as they did, by the inverse. Thanks. – DonAntonio Dec 25 '12 at 2:53

The reason that the action $\sigma:(v_1,\cdots,v_n)\mapsto(v_{\sigma(1)},\cdots,v_{\sigma(n)})$ is not left, is because when you apply a second permutation element $\rho$ to $\sigma v$, $v_{\sigma(1)}$ is the first coordinate of $\sigma v$, not the $\sigma(1)$th, $v_{\sigma(2)}$ is the second coordinate of $\sigma v$, not the $\sigma(2)$th, etc. so we cannot simply apply $\rho$ to the subscript in the seemingly obvious (but naive) way. Instead, we would write $w:=\sigma v=(v_{\sigma(1)},\cdots,v_{\sigma(n)})$, so that we have $w_i=v_{\sigma(i)}$, in which case $w_{\rho(i)}=v_{\sigma(\rho(i))}$ hence $\rho(\sigma v)=(v_{\sigma\rho(1)},\cdots,v_{\sigma\rho(n)})=(\sigma\rho)v$, in which case it is a right action (and it makes more sense to write it as group-element-on-the-right instead).

I fill in Ted's exquisite answer.

This is tricky - the two cases look the same, but they're not.
The first one is a right action $v \cdot (p_1 \cdot p_2) = (v \cdot p_1) \cdot p_2$,
while the second one is a left action $p_1 \cdot (p_2 \cdot f) = (p_1 \cdot p_2) \cdot f$.
To see why, consider these 2 permutations:

$\color{tomato}{p_1(1) = 1, p_1(2) = 3, p_1(3) = 2}$

$\color{tomato}{p_2(1) = 3, p_2(2) = 2, p_2(3) = 1}.$

Let's write out explicitly what the actions are in the two cases to see the difference.

First, let's work out what the compositions $p_1 \cdot p_2$ and $p_2 \cdot p_1$ are.

The composition $p_1 \cdot p_2$ is: $\qquad p_1(p_2(1)) = 2, p_1(p_2(2)) = 3, p_1(p_2(3)) = 1$

while the composition $p_2 \cdot p_1$ is: $\qquad p_2(p_1(1)) = 3, p_2(p_1(2)) = 1, p_2(p_1(3)) = 2.$

Observe that they are not the same. We will use these later.

Now let's look at the 2 actions. The first action is on vectors. By the definition of the first action, $$p_1 \cdot (v_1, \color{magenta}{v_2}, \color{limegreen}{v_3}) = (v_{\LARGE{p_1(1)}}, \color{limegreen}{v_{{\LARGE{p_1(2)}}}}, \color{magenta}{v_{\LARGE{p_1(3)}}}) = (v_1, \color{limegreen}{v_3}, \color{magenta}{v_2}).$$

In words: $p_1$ acting on a vector interchanges the second and third coordinates. Similarly, $$p_2 \cdot (v_1, v_2, v_3) = (v_3, v_2, v_1). \tag{♥}$$ In words: $p_2$ acting on a vector interchanges the first and third coordinates.

In this situation, I find that thinking in words reduces the confusion: $p_1$ interchanges the second and third coordinates, not $v_2$ and $v_3$. You'll see the difference below. Hence:

\begin{align} p_1 \cdot (p_2 \cdot v) = p_1 \cdot (p_2 \cdot (v_1, v_2, v_3)) = by (♥) = p_1 \cdot (v_3, \color{magenta}{v_2}, \color{limegreen}{v_1}) & = (v_3, \color{limegreen}{v_1}, \color{magenta}{v_2}) \tag{☼}\\ & \neq (v_{\LARGE{p_1(1)}}, v_{{\LARGE{p_1(2)}}}, v_{\LARGE{p_1(3)}}). \end{align}

If the last equality seems wrong, use the "words" description of $p_1$: Interchange the second and third coordinates.)

Question: Does $p_1 \cdot (p_2 \cdot v) = (p_2 \cdot p_1) \cdot v$ or $(p_1 \cdot p_2) \cdot v$? Don't need to calculate these two.
Just look at the orange to determine if $(v_3, \color{limegreen}{v_1}, \color{magenta}{v_2})$ in $(☼)$ is $(v_{\Large{p_1(p_2)(1)}},v_{\Large{p_1(p_2)(2)}}, ...)$ or $(v_{\Large{p_2(p_1)(1)}}, v_{\Large{p_2(p_1)(2)}}, ...)$.

Conclusion: For the first action, on vectors, $p_1 \cdot (p_2 \cdot v) = (p_2 \cdot p_1) \cdot v$.
So this is not a left action, but a right action.

Now look at the second action, which is not on vectors, but on real-valued functions on the set of all vectors. By definition, $$(p_1 \cdot f)(x_1, x_2, x_3) = f(x_{\LARGE{p_1(1)}}, x_{\LARGE{p_1(2)}}, x_{\LARGE{p_1(3)}}) = f(x_1, x_3, x_2).$$

In words: To evaluate $p_1$ applied to a function at a vector, interchange the second and third coordinates of the vector, then apply the function.

Similarly, for $p_2$, $(p_2 \cdot f)(\color{limegreen}{x_1}, x_2, \color{magenta}{x_3}) = f(\color{magenta}{x_{\LARGE{p_2(1)}}}, x_{\LARGE{p_2(2)}}, \color{limegreen}{x_{\LARGE{p_2(3)}}}) = f(\color{magenta}{x_3}, x_2, \color{limegreen}{x_1})$

The words version is: To evaluate $p_2$ applied to a function at a vector, interchange the first and third coordinates of the vector, then apply the function.

So what is $p_1 \cdot (p_2 \cdot f)$, applied to a vector? It is \begin{align} (p_1 \cdot (p_2 \cdot f))(x_1, x_2, x_3) = (p_2 \cdot f)(\color{limegreen}{x_1}, x_3, \color{magenta}{x_2}) & = f(\color{magenta}{x_2}, x_3, \color{limegreen}{x_1}) \\ & \neq f(x_{\LARGE{p_2(1)}}, v_{{\LARGE{p_2(3)}}}, v_{\LARGE{p_2(2)}}). \end{align}

(Again, using the "words" description may reduce the confusion.) Now is this last term $p_1 \cdot p_2$ or $p_2 \cdot p_1$ applied to $f$? It is the former, as you can see from the orange calculations of $p_1 \cdot p_2$ above.

Conclusion: $p_1 \cdot (p_2 \cdot f) = (p_1 \cdot p_2) \cdot f$. This one is a left action.

I hope this clears up why these two cases aren't the same. Of course, I haven't yet proven that these are actions in general (I've only illustrated it for the particular $p_1$ and $p_2$ above), but hopefully this will give you the right idea for the general proof.

The key point is to remember that with these definitions, we are permuting the coordinates of the vector according to $p$, rather than the indices of the $v$'s according to $p$. If we did the latter instead, then the left and right action cases above would be reversed. See alias vs alibi for more discussion of this point.