The Lie algebra of a closed subgroup is a linear subspace
Statement
Let be a closed subgroup and
Show that is a linear subspace of .
Hint
Closure under scalar multiples is easy. For sums, use the Lie product formula applied to and , together with the closedness of .
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Solution
Idea. The sum is approximated through products of elements of , and is closed.
Contains . , so .
Scalar multiples. If and , then for every : because is just another real number. So .
Sums. Let and fix . For every integer , and ( contains for every real ), hence the product
By the Lie product formula applied to , , as . The limit is invertible (it is an exponential), so it lies in , and is closed in : . This holds for every , so .
Remark. The hypothesis "closed" is used in the last step. For the non-closed subgroup of the torus in the limit need not stay in the subgroup: its closure is the whole torus, whose Lie algebra is two-dimensional, larger than the line generated by the winding direction.