A representation of a Lie group $G$ on a vector space $V$ is defined to be a Lie group homomorphism $\rho$ such that $$\rho: G \rightarrow GL(V),$$ whereas the representation of a Lie algebra $\mathfrak{g}$ on a vector space $V$ is defined to be a Lie algebra homomorphism $\phi$ such that $$\phi: \mathfrak{g} \rightarrow \text{End}(\phi).$$

Why is the image space for a Lie algebra representation less restricted than for a Lie group (requiring only a homomorphism instead of an isomorphism)?

  • $\begingroup$ Because the Lie algebra of general linear groups are matrix groups. $\endgroup$
    – Alexey Do
    Commented Jun 7, 2023 at 4:44
  • $\begingroup$ @AlexeyDo I'm sorry, could you explain how that forces the image of $\phi$ to be an endomorphism? $\endgroup$
    – CBBAM
    Commented Jun 7, 2023 at 4:46
  • 1
    $\begingroup$ The calculation you need to make is to show that the Lie algebra $\mathfrak{gl}(V)$ of $GL(V)$ (for $V$ either a real or complex vector space) is $\text{End}(V)$ (equipped with the commutator bracket). What we actually care about is this Lie algebra $\mathfrak{gl}(V)$. This setup guarantees that the derivative of a Lie group representation is a Lie algebra representation. $\endgroup$ Commented Jun 7, 2023 at 4:54
  • $\begingroup$ @QiaochuYuan Because the Lie algebra of $GL(V)$ is $\text{End}(V) = M_{n\times n}$, we naturally define $\phi$ as it is in my post so that the derivative of $\rho$ induces $\phi$? $\endgroup$
    – CBBAM
    Commented Jun 7, 2023 at 5:00
  • 3
    $\begingroup$ Note for example that $0\in\mathfrak{g}$ cannot ever represent an automorphism. A group contains only invertible elements and automorphisms are invertible. While a Lie algebra is linear and so its representation should live in a linear space. Said another way, a Lie group rep is a group homomorphism while a Lie algebra rep is a Lie algebra homomorphism $\endgroup$
    – Callum
    Commented Jun 7, 2023 at 7:23

3 Answers 3


The short answer: Because $GL(V)$ is a group and $\mathrm{End}(V)$ is a Lie algebra.

In particular, $GL(V)$ has the following:

  • Every element has an inverse
  • There is no addition of matrices, only multiplication

Whereas $\mathrm{End}(V)$ has the following:

  • It is a vector space, i.e. there is a $0$ element, we can scale matrices with scalars and we can add matrices.
  • It has a funny multiplication $[X,Y] = XY-YX$ making it into an algebra.

So it is only natural to represent a Lie group in the group $GL(V)$, and a Lie algebra in the algebra $\mathrm{End}(V)$.


It might also be useful to note that if $(V,\rho)$ is a representation of a Lie group $G$, then the elements of $G$ are mapped to automorphisms of $V$, but to obtain the associated Lie algebra homomorphism, you differentiate $\rho$ at the identity element: $\phi=d\rho\colon T_eG\to T_e\mathrm{GL}(V)$ and so one answer to your question is that the reason you get an endomorphism is because the vector space of all endomorphisms is what you get for the tangent space at the identity of $\mathrm{GL}(V)$.

The fact that $T_e\mathrm{GL}(V)=\mathrm{End}(V)$ is easy, it follows immediately from the fact that $\mathrm{GL}(V)$ is open in $\mathrm{End}(V)$, and hence if you start at $I_V$ the identity linear map in $\mathrm{GL}(V)$, then you can move a little bit at least in any direction you like in $\mathrm{End}(V)$. (The fact that $\mathrm{GL}(V)$ is open is also easy -- for example, it follows from the fact that $\mathrm{GL}(V)$ is the complement of the hypersurface $\{A \in \mathrm{End}(V): \mathrm{det}(A)=0\}$.)

It may also be useful to remember that the Lie algebra structure on $T_eG$ comes from identifying $T_eG$ with the Lie algebra of left-invariant vector fields or derivations on $G$, which is a sub-Lie algebra of the Lie algebra of all left-invariant derivations. It follows from the definitions that $\phi = d\rho$ is a Lie algebra homomorphism from left-invariant derivations on $G$ to left-invariant derivations on $\mathrm{GL}(V)$. In that sense, it is not ``just'' a map to $\mathrm{End}(V)$ the space of linear self-maps of $V$, it is a map on left-invariant derivations, and derivations are the infinitesimal analogue of automorphisms.

Up to the question of the centre, which is the kernel of the adjoint representation, you can pretty much see this through the map $\phi$: indeed if $X \in \mathfrak g$ then if we set $a_X = \mathrm{ad}(\phi(X))$ then $a_X$ is a derivation of $\mathrm{End}(V)$, that is, its action on products obeys the Leibniz product rule: $$ \begin{split} a_X(Y_1)Y_2 + Y_1a_X(Y_2) &= (XY_1-Y_1X)Y_2 + Y_1(XY_2-Y_2X) \\ &= XY_1Y_2 - Y_1XY_2 + Y_1 XY_2 + Y_1Y_2X \\ &= X(Y_1Y_2) - (Y_1Y_2)X = a_X(Y_1Y_2) \end{split} $$ so that $\phi$ induces a map $a\colon \mathfrak g \to \mathrm{Der}(\mathrm{End}(V))$ where $$ \mathrm{Der}(\mathrm{End}(V))= \{\alpha \in \mathrm{End}(\mathrm{End}(V)): \alpha(Y_1Y_2) = \alpha(Y_1)Y_2 + Y_1 \alpha(Y_2), \forall Y_1,Y_2 \in \mathrm{End}(V)\}. $$


Representations in very broad sense associate (through a homomorphism) for every element of abstract object an element of specimen objects (they are specimens of the same kind as the abstract objects).

For example when the abstract objects are groups, we talk of representation theory of groups: then speciment Objects in this case are $GL(V)$ for various vector spaces $V$ (over various fields).

Now for for an abstract Lie Algebras (over a field $k$) , the SPECIMENS are the very natural concrete examples, viz., $End(V)$ for various $V$ on the same field $k$. $Aut(V)$ DOES NOT satisfy the axioms of a Lie Algebra only End(V) does.

  • $\begingroup$ To the people who downvoted my answer: I'd be glad to know if I have misunderstood the question or anything wrong in my answer. Both me and the poser of the question can learn something from you. $\endgroup$ Commented Jun 8, 2023 at 3:17
  • $\begingroup$ I myself did not downvote, but I strongly suspect that some people are seriously put off by the all-caps instances... Just some version of "italic" would be more acceptable, by current standards, I guess. $\endgroup$ Commented Jun 24, 2023 at 18:56
  • $\begingroup$ @paulgarrett: Thanks for taking efforts to reply. I see the point in your reply and made the change getting rid of that irritant in my answer. $\endgroup$ Commented Jun 26, 2023 at 3:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .