Representation vs group action?

Question

I'm working this problem from Howe's An Invitation to Representation Theory:

Let $G = (\mathbb{R}^*, \times)$ be the group of non-zero real numbers under multiplication, and let $P_2$ denote the vector space of polynomials (in one variable) of degree two or less. Verify that the action of $G$ on $P_2$ given by $a\mathbin{.}p(x) = p(a^{-1}x)$ is a representation of $G$ on $P_2$.

I'm not sure where to start because I'm having trouble differentiating the group action from the representation. Any advice?

Answer 1

A representation of a group $G$ on a $\mathbb{k}$-vector space $V$ is a homomorphism $$ \rho: G \to \operatorname{GL}(V), $$

Explicitly, the conditions that make this map a homomorphism are $$ \rho(1)(v) = v \quad\text{and}\quad \rho(gh)(v) = \bigl(\rho(g) \circ \rho(h)\bigr) (v). \tag{A} $$ for all $g, h \in G$ and all $v \in V$ or in coordinates,

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In your example, $G = \mathbb{R}^*$ and $V = P_2 = P_2(\mathbb{R})$ and we can check that the proposed map $\rho: \mathbb{R}^* \to \operatorname{GL}(P_2)$, given by $a \mapsto \rho_a$, where $\rho_a: P_2 \to P_2$ for each $a \in \mathbb{R}^*$ is the scaling transformation $p(x) \mapsto p(a^{-1}x)$, satisfies the required properties:

(0) $\rho_a$ is linear and has appropriate source and target. Certainly, it's well-defined since $a \neq 0$, and we can verify that $p(a^{-1}x)$ is still a polynomial in $x$ of degree at most $2$. Then you have to check linearity.

(1) $\rho_a$ is a group action. These are the properties in $(A)$. The only potential subtlety is with the second property:

$$ \rho_{ab} \bigl(p(x)\bigr) = p\bigl((ab)^{-1} x) = p\bigl(b^{-1}a^{-1} x\bigr) = \rho_b \bigl( p(a^{-1} x)\bigr) = \rho_a \circ \rho_b \bigl(p(x)\bigr). $$

Since $G$ is abelian, writing $(ab)^{-1}$ as $b^{-1}a^{-1}$ might seem superfluous, but it's what makes the calculation work, which also reveals that it's the right kind of formula for similar situations where the group isn't abelian.

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By the way, if we have a basis $\{v_1, \dots, v_n\}$ chosen for $V$, we can do this all in coordinates. The representation is equivalent to a homomorphism $$ \rho: G \to \operatorname{GL}_n(\mathbb{k}), $$ producing invertible $n \times n$ matrices for each $g \in G$. Let's use the notation $\rho_g := \rho(g)$, so we require the matrices to satisfy $$ \rho_1 = I_n \quad\text{and}\quad \rho_{gh} = \rho_g\,\rho_h. $$

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We can also work out the example in coordinates since $P_2$ has a nice basis given by monomials: $\{1, x, x^2\}$. We need to construct for each $a \in \mathbb{R}^*$, the $3 \times 3$ matrix representing pre-multiplication by $a^{-1}$. The action on each basis vector is

$$ \rho_a(1) = 1, \quad \rho_a(x) = a^{-1}x, \quad\text{and}\quad \rho_a(x^2) = (a^{-1}x)^2 = a^{-2}x^2 $$

hence for each $a \in \mathbb{R}^*$, the matrix $\rho_a$ is

$$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & a^{-1} & 0 \\ 0 & 0 & a^{-2} \end{bmatrix} $$

and we can easily check that these matrices satisfy the required identities.

Answer 2

Beeing a representation is more restrictive than beeing an action. A representation always comes from an action, but not every action is a representation. A representation of a group $G$ is an action defined on a vector space $V$ such that the maps on $V$ defined by the elements of $G$ are linear.

For example, the group of the bijections $\mathcal S(\mathbb R)$ from $\mathbb R$ to itself defines a natural action on the vector space $\mathbb R$ by $$ \forall f \in \mathcal S(\mathbb R), \forall x \in \mathbb R, f \cdot x := f(x). $$ This action is not a representation, since it is not linear (for example, $x \mapsto x^3$ is a non-linear bijection of $\mathbb R$).

A representation of a group is an action on a vector space that is linear. So the group of the linear bijections $\mathrm{GL}(\mathbb R)$ from $\mathbb R$ to itself (which is isomorphic to $\mathbb R^{*}$) defines a natural action on the vector space $\mathbb R$ by

$$\forall f \in \mathrm{GL}(\mathbb R), \forall x \in \mathbb R, f \cdot x := f(x) $$

and this action is also a representation since the maps $f$ are linear.

So in your example you have to check two facts :

1. That the relation $a \cdot p(x) := p(a^{-1}x)$ is indeed an action.
2. That for every $a \in G$, the map $x \in P_2 \mapsto p(a^{-1}x)$ is linear, which will show that this action is a representation of $G$.