Why $SL_{n}(k)$ acts on $V$ by restriction?

Question

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.

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.