Bending Representations

Here's a very well-studied construction that I think everyone should know, called bending representations.  I'll start with the algebraic picture, and then get into the relationship with topology.  (At the end we'll hopefully prove that a splice of two knots in the three-sphere has positive dimensional $SU(2)$ character variety.)  

 First, recall that given a group $G$, an $SU(2)$ representation is just a homomorphism $\rho: G \to SU(2)$.  The representation variety is $R(G) = Hom(G,SU(2))$.  Now $SU(2)$ acts on $R(G)$ by conjugation: for $A \in SU(2)$, define $\rho^A(g) = A \rho(g) A^{-1}$.  We define the $SU(2)$ character variety of $G$, $\chi(G)$, to be the quotient of $R(G)$ by conjugation.  Note that $\rho^{\pm Id} = \rho$. A representation $\rho$ is irreducible if $\rho^A = \rho$ implies $A = \pm I$.  For some motivation to those topologically inclined, one of the Kirby problems is to prove that every homology three-sphere other than $S^3$ admits an irreducible $SU(2)$ representation.   On the other hand, if $\rho$ has abelian image, then $\rho$ is not irreducible; indeed every non-trivial abelian subgroup of $SU(2)$ is $\{\pm I\}$ or a conjugate of $S^1$, i.e. the subgroup of diagonal matrices.  If the image of $\rho$ is $\{ \pm I\}$, then the stabilizer of $\rho$ is all of $SU(2)$; if it is a conjugate of $S^1$, then that subgroup will be the stabilizer.  

Now, suppose you have two representations $\rho: G \to SU(2)$ and $\rho': G' \to SU(2)$.  Then you can build a new representation out of $\rho, \rho'$, such as $\rho* \rho' : G * G' \to SU(2)$.  However, you can build lots of other representations.  For each $A \in SU(2)$, then we can also define a new $SU(2)$-representation of $G * G'$ by $\rho^A * \rho'$.  If $\rho, \rho'$ are irreducible, then $\rho^A * \rho'$ will not be conjugate to $\rho^B * \rho'$ for any $A \neq \pm B$.  As a result, if $G$, $G'$ admit irreducible representations, then the character variety of $G * G'$ has a positive dimensional component by varying over $A$ in $SU(2)/\pm I$.  (For context, it can happen that $G$ has irreducible representations, but $\chi(G)$ is 0-dimensional.  I.e. the only way to continuously deform a representation is to conjugate.  This happens for the fundamental groups of the Brieskorn spheres $\Sigma(p,q,r)$ for example.)  

We can do a slightly fancier version of this.  Suppose $H$ is a subgroup that embeds in $G$ and $G'$.  Then if $\rho: G \to SU(2)$ and $\rho': G' \to SU(2)$ agree on $H$, then by the universal property of amalgamated products, we get a representation $\rho *_H \rho'$ of $G *_H G'$.  (Basically, for a word in the amalgamated product, do $\rho$ on $G$-letters and $\rho'$ on $G'$-letters, and since they agree on $H$-letters, this is well-defined.)  

 Let's now additionally assume $H$ is abelian (e.g. think $G, G'$ are the fundamental groups of knot complements and $H$ is the peripheral subgroup, generated by the meridian and longitude and isomorphic to $\mathbb{Z}^2$.)  In this case, there's a subgroup $\Gamma \subset SU(2)$ conjugate to $S^1$ such that $\rho|_H$ is stabilized by $\Gamma$.  (This is because $\rho(H)$ is an abelian subgroup of $SU(2)$, and so it's stabilized by either $SU(2)$ or a subgroup conjugate to $S^1$.)  That means that $\rho^A *_H \rho'$ is a representation of $G *_H G'$.  Now, if $\rho$ is irreducible, then $\rho$ is not stabilized by $\Gamma$ even though $\rho|_H$ is.  If $\rho'$ is also irreducible, then $\rho^A *_H \rho'$ will not be conjugate to $\rho^B *_H \rho'$ for $A \neq \pm B$ with $A, B \in \Gamma$.  Hence, $\chi(G *_H G')$ has dimension at least 1, since $\Gamma$ is 1-dimensional.  This is a good example of bending $\rho *_H \rho'$ to produce lots of new representations of this interesting group. 

Who cares?  Suppose that $K, K'$ are non-trivial knots in $S^3$.  In this paper, Zentner uses gauge theory to prove that there exist irreducible representations $\rho : \pi_1(S^3 - K) \to SU(2)$ and $\rho': \pi_1(S^3 - K') \to SU(2)$ such that $\rho(\mu_K) = \rho'(\lambda_{K'})$ and $\rho(\lambda_K) = \rho'(\mu_{K'})$.  If $Y(K,K')$ denotes the splice of $K$ and $K'$, i.e. glue the exteriors of $K$ and $K'$ together by exchanging meridian and longitude, then a representation $\rho$ of $\pi_1(S^3 - K)$ and $\rho'$ of $\pi_1(S^3 - K')$ such that $\rho(\mu_K) = \rho'(\lambda_{K'})$ and $\rho(\lambda_K) = \rho'(\mu_{K'})$ induces a representation of $\pi_1(Y(K,K')) = \pi_1(S^3 - K) *_{\mathbb{Z}^2} \pi_1(S^3 - K')$.  By bending, we get that the character variety of $Y(K,K')$ has positive dimension!  (This is cool!  For example, it shows us that the dimension of the character variety knows the difference between the Brieskorn spheres $\Sigma(p,q,r)$ and knot splices!)

Here's an even cooler converse to our discussion.  Say $Y$ is a homology sphere (more restrictive than necessary), and $\dim \chi(Y) > 0$, then $Y$ must contain an essential surface!  This is a consequence of Culler-Shalen theory.  (For more details, see these notes.)  

In general, the dimension of the representation or character variety can contain a lot of useful information, especially if one works with more general Lie groups.  For example, this is a key part of Agol's proof that ribbon concordance is a partial order.