My professor defined the meaning of a group action by $G$(a group) on $X$(a set) is a function:

$$G \times X \rightarrow X$$defined by $$(g, x) \mapsto {}^gx$$ Such that $${}^g({}^hx) = {}^{(gh)}x$$ for all $g,h \in G.$

Then he gave us examples for this group action definition which are:

1- $GL_{n}(V)$(where $V$ is a vector space over $k$) acts on $V$ by linear transformation.

2- $SL_{n}(k)$ acts on $V$ by restriction.

My questions are:

1- does not $GL_{n}(V)$ acts on $V$ by invertiblelinear transformation?

2- What is the meaning of " $SL_{n}(k)$ acts on $V$ by restriction"? restriction of what?

Note: I know that $SL_{n}(k)$ is the special linear group of degree $n$ over $k$ which is the group of matrices of determinant 1.

  • $\begingroup$ wrt q1: Is it possible that the general linear group elements' invertibility is independent of the group action's invertibility? $\endgroup$
    – Dohleman
    Oct 27, 2020 at 0:12
  • $\begingroup$ @Dohleman could you explain more please? $\endgroup$
    – user778657
    Oct 27, 2020 at 0:15
  • $\begingroup$ a requirement of the identity linear transformation for the general linear group is that it has certain kernel and range properties. how can you use this to look at the structure of the group action and talk about that in terms of potential invertibility? $\endgroup$
    – Dohleman
    Oct 27, 2020 at 0:28

1 Answer 1


Note that is $GL(k)$ consists by definition of the invertible linear transformations on $V$. This set forms a group (Identity transformation is invertible, composition of invertible is invertible, etc). The set of all linear transformations on V does not form a group (it does form a ring).

The meaning of restriction here is the following. If the group $G$ acts on the set $X$ and $H$ is any subgroup of $G$ then $H$ acts on $X$ by restricting the action map from $G\times X$ to $H\times X$. In other words, you just use the formula but apply it to elements of $H$ only.

This is precisely the situation with the subgroup $SL(V)$ - it acts on $V$ via the same action map, namely matrix multiplication.

  • 1
    $\begingroup$ To clarify part 1 - elements of GL are invertible linear maps, hence by definition act on V by an invertible map (corresponding to that element). Note that any group action must be by invertible bijections however. $\endgroup$
    – the_lar
    Oct 27, 2020 at 0:21
  • $\begingroup$ So you are saying that I am correct for part 1 ..... correct? $\endgroup$
    – user778657
    Oct 27, 2020 at 3:18
  • $\begingroup$ In part one, part of the definition of a ring is to be an Abelian group. $\endgroup$
    – user778657
    Oct 27, 2020 at 3:19
  • $\begingroup$ Yes, you are correct in part 1, it’s almost a tautology. $\endgroup$
    – the_lar
    Oct 27, 2020 at 4:25
  • $\begingroup$ The set of all linear transformations is a ring with the two operations T+S and TS. The second one (composition) is the one you use to get a group. The first one (addition) gives you a vector space structure. $\endgroup$
    – the_lar
    Oct 27, 2020 at 4:27

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