Harish-Chandra module

Remarks

The derivative in clause 2 makes sense by clause 1, since it is taken inside a finite-dimensional space.

Example: let $\pi$ be a finite-dimensional continuous representation of $G$ on $V$, with differential $d\pi$ extended complex-linearly to $\mathfrak g$. Then $V$ with $X \cdot v = d\pi(X) v$ and $\pi|_K$ is a Harish-Chandra module. Clause 1 holds because $V$ itself is finite-dimensional, clauses 2 and 3 hold by the definition of $d\pi$ and by $\pi(k) \pi(\exp tX) \pi(k)^{-1} = \pi(\exp t \mathrm{Ad}(k) X)$, and clauses 4 and 5 hold because $\dim V < \infty$.

Non-example: an infinite direct sum of copies of the trivial one-dimensional module, with $\mathfrak g$ acting by $0$ and each $k \in K$ acting as the identity, is a $(\mathfrak g, K)$-module. It is not a Harish-Chandra module, because the trivial representation of $K$ occurs infinitely often and clause 5 fails.

If $\pi$ is an admissible representation of $G$ on a Hilbert space $H$, its $K$-finite vectors are smooth and form a Harish-Chandra module when $\pi$ has finite length. Two such representations are called infinitesimally equivalent when their Harish-Chandra modules are isomorphic.