What is the difference between a map being linear in linear algebra and a map being linear representation in linear representation theory?

I know from the answers at the back of the book that the following map:

$$(S(t)f)(x) = f (tx),$$ where $S: R \rightarrow S(V)$ and $V$ is the subspace of all polynomials with real coefficients and $t \in \mathbb{R}, f \in V.$

Is not a linear representation (but I do not know how?). Even though I found it is linear when I put instead of $f$, $\alpha f + \beta f$. I know that for the given map to be a linear representation it must be a homomorphism (I am not very sure from this information, is it correct?) so what are the operations that I should consider here for studying homomorphism?

Could anyone clarify this discrepancies for me please?

  • $\begingroup$ What is $R$ in this case? What is $S(V)$? $\endgroup$ – Omnomnomnom Jan 12 at 17:19
  • $\begingroup$ R is the reals .... S(V) the group of something(which I do not know) over the vector space V @Omnomnomnom $\endgroup$ – hopefully Jan 12 at 17:25
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    $\begingroup$ Which book are you using? $\endgroup$ – user458276 Jan 12 at 17:36
  • $\begingroup$ @user458276 "Linear Representations of groups" for Ernest B. Vinberg $\endgroup$ – hopefully Jan 12 at 17:56

For all $t$, $S(t)$ is indeed a linear map; but $t\mapsto S(t)$ is not a morphism, because $S(t+t') \neq S(t)\circ S(t')$ in general.

If you consider $\mathbb{R}^*\to GL(\mathbb{R}[X])$, $t\mapsto S(t)$, it will, however be a morphism because $(S(tt')f )(x) = f(tt'x) = f(t\cdot (t'x))= S(t)f(t'x) = S(t)(S(t')f)(x) = (S(t)\circ S(t') (f))(x)$ so $S(tt') = S(t)\circ S(t')$.

A linear representation of a group $G$ is a morphism $G\to GL(V)$, that is, for each $g\in G$ you have a linear map $\rho_g : V\to V$ (subject to certain conditions). $\rho$ is a group morphism, $\rho_g$ is linear.

Here $S(t)$ is linear but $t\mapsto S(t)$ is not a group morphism from $(\mathbb{R},+)$

  • $\begingroup$ What is S(V) here and why $S(t + t^')$$ \neq $$S(t) \circ S(t^')$, could you give me an example please? $\endgroup$ – hopefully Jan 12 at 18:01
  • $\begingroup$ Why "$S(t+t') \neq S(t)\circ S(t')$ in general" , could you give me an example please? $\endgroup$ – hopefully Jan 12 at 18:08
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    $\begingroup$ Please think about it and try to see what it would mean for $S(t+t')$ to be $S(t)\circ S(t')$. Pick almost any example of polynomial to see $\endgroup$ – Max Jan 12 at 18:17
  • $\begingroup$ okay thank you so much :) $\endgroup$ – hopefully Jan 12 at 18:22
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    $\begingroup$ It's probably the group of invertible linear operatirs (but not $n\times n$ because $V$ is infinite dimensional) $\endgroup$ – Max Jan 12 at 18:31

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