There is a deep famous correspondence between analytic and algebraic properties.
For a complex curve $X$, representations $Hom(\pi_1(X), U(n))$ correspond to degree $0$ semistable bundles. This is a hard theorem but one direction is easy; given a representation you can easily build the bundle with flat connection and holomorphic metric and thus see it's semistable.
It turns out that representations $Hom(\pi_1(X), GL(n))$ correspond to semistable Higgs bundle.
I want whatever easy part of this correspondence you can tell me; given a representation we can build a vector bundle with a flat connection, but how do we build the map $V \to V \otimes(K)$? Note that $H^0(V,V \otimes(K)) = H^1(End(V))^*$ is the cotangent space of $V \in Bund(X)$.
From this story, we see that we expect there to be a map from the cotangent space of representations $T^*Hom(\pi_1,U(n)) \to Hom(\pi_1(X), GL(n))$! For a representation $\rho$, we can describe the former cotangent space as $$H^1(\pi_1 , Ad(\rho)$$. Given $e \in H^1(\pi_1 , Ad(\rho)$, how do we get an element of $Hom(\pi_1(X), GL(n))$?
Warning; I've heard that these equivalent spaces have different complex structure, so maybe what I'm asking is unreasonable because the cotangent spaces aren't canonically identified?
One direction which makes the theorem vaguely make sense is:
Given a representation $\rho:\pi_1(X) \to GL_n$ we get a bundle $V$ with a connection $\nabla$. Let's suppose this $V$ is stable, so that we can find a unique unitary representation $\rho'$ giving a flat unitary connection $\nabla'$ (Whatever that means). Then $\nabla - \nabla'$ is the desired twisted endomorphism $V \to V \otimes K$! Thus if we believe the (hard) theorem that says semistable bundles come from unitary representations we can easily define one direction of the Higgs bundle.