# Non-surjectivity of exponential map: how to understand?

I'm given to understand the exponential map is not generally surjective -- the standard example is $$\mathrm{SL}(\mathbb{R}^2)$$ [ 1 ].

I can clearly see why this is so in the non-connected case -- the tangent space is a tangent space to the connected component alone, so its image must be contained in the connected component. I do not see why the map isn't surjective in the connected case.

I also don't see why the map is then again surjective in the compact case -- wikipedia claims that this is a special case of "the exponential map is surjective if every element is contained in a maximal torus". Is this right? Is there a good way to understand why this is true?

Note that I am not looking for counter-examples: I'm aware of them. I'm looking for intuition -- perhaps a clever look at what the image of the exponential map actually looks like in the non-surjective case (how it "misses" some of the points in the group).

As an analogy, if asked to explain smooth non-analytic functions, it would be more instructive (than simply providing the example of $$e^{-1/x}$$) to explain that a function may grow slower than all polynomials near zero -- and provide the construction as $$1/f(1/x)$$ from any function $$f$$ that grows faster than all polynomials as $$x\to\infty$$.

(See here for more examples of the kind of intuition I'm looking for, within the context of Lie theory.)

• The exponential map is surjective around $1$ but its image is not a subgroup. For $GL(n,\Bbb{R})$ one frontier is the radius of convergence of $\log(I+.)$ and its image under conjugaison. It is surjective iff every element is $=\exp(A)$ thus contained in a torus $\exp(tA)$ – reuns Oct 7 '19 at 1:03
• There's a standard proof for the compact case. In summary: (1) Since $G$ is compact, it admits a bi-invariant metric. (2) For a bi-invariant metric the (Lie algebra) exponential and the (geodesic) exponential maps coincide. (3) Apply the Hopf-Rinow Theorem. terrytao.wordpress.com/2011/06/25/… – Travis Willse Oct 7 '19 at 4:52