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There is a topological criterion due to Thurston. Using the JSJ machine (and work of many others) this criterion can also be phrased algebraically. I'll essay these below. Please note that the situation is much simpler for knots. To answer your question most directly, here is the desired reference to Wikipedia.

http://en.wikipedia.org/wiki/Hyperbolic_link

This page refers to the books of Colin Adams and William Thurston. Both are excellent.

Now, here is Thurston's criterion.

Suppose that $L$ is the link and $X$ is the link complement. Suppose $\pi = \pi_1(X)$. We assume the following.  $\newcommand{\ZZ}{\mathbb{Z}}$

• $L$ is not a split link. Equivalently, $X$ is irreducible. Equivalently, $\pi$ is not a free product.
• $L$ is not the unknot. Equivalently (as $L$ is not split) $\partial X$ is incompressible. Equivalently, $\pi$ is not $\ZZ$.
• $L$ has no cabled component, and has no parallel components. Equivalently, $X$ is acylinderical.
• $L$ has no component that is an "undisturbed satellite". Equivalently, $X$ is atoroidal. These last two topological properties are equivalent to $\pi$ not containing a copy of $\ZZ^2$.

Then $X$ admits a hyperbolic structure.

There is a topological criterion due to Thurston. Using the JSJ machine (and work of many others) this criterion can also be phrased algebraically. I'll essay these below. Please note that the situation is much simpler for knots. To answer your question most directly, here is the desired reference to Wikipedia.

http://en.wikipedia.org/wiki/Hyperbolic_link

This page refers to the books of Colin Adams and William Thurston. Both are excellent.

Now, here is Thurston's criterion. (EDIT: exposition improved after reading Bruno Martelli's answer.)

Suppose that $L$ is the link and $X$ is the link complement. Suppose $\pi = \pi_1(X)$. We assume the following properties (and each property assumes the proceeding ones).  $\newcommand{\ZZ}{\mathbb{Z}}$

• $L$ is not a split link. Equivalently, $X$ is contains no essential two-sphere. Equivalently, $\pi$ is not a free product.
• $L$ is not the unknot. Equivalently, $X$ contains no essential disk. Equivalently, $\pi$ is not $\ZZ$.
• $L$ has no component that is an "undisturbed satellite knot". Equivalently, $X$ contains no essential torus.
• $L$ is not a torus knot. Equivalently, $X$ contains no essential annulus. These last two topological properties are equivalent to $\pi$ not containing a copy of $\ZZ^2$.

Then $X$ admits a hyperbolic structure.

There is a topological criterion due to Thurston. Using the JSJ machine (and work of many others) this criterion can also be phrased algebraically.

Suppose that $L$ is the link and $X$ is the link complement. Suppose $\pi = \pi_1(X)$. We assume the following. $\newcommand{\ZZ}{\mathbb{Z}}$

• $L$ is not a split link. Equivalently, $X$ is irreducible. Equivalently, $\pi$ is not a free product.
• $L$ is not the unknot. Equivalently (as $L$ is not split) $\partial X$ is incompressible. Equivalently, $\pi$ is not $\ZZ$.
• $L$ has no cabled component, and has no parallel components. Equivalently, $X$ is acylinderical.
• $L$ has no component that is an "undisturbed satellite". Equivalently, $X$ is atoroidal. These last two topological properties are equivalent to $\pi$ not containing a copy of $\ZZ^2$.

Then $X$ admits a hyperbolic structure as you describe. This is just part of a much bigger story, of course. Let me point you towards Wikipedia and the references therein.

http://en.wikipedia.org/wiki/Hyperbolic_link

There is a topological criterion due to Thurston. Using the JSJ machine (and work of many others) this criterion can also be phrased algebraically. I'll essay these below. Please note that the situation is much simpler for knots. To answer your question most directly, here is the desired reference to Wikipedia.

http://en.wikipedia.org/wiki/Hyperbolic_link

This page refers to the books of Colin Adams and William Thurston. Both are excellent.

Now, here is Thurston's criterion.

Suppose that $L$ is the link and $X$ is the link complement. Suppose $\pi = \pi_1(X)$. We assume the following. $\newcommand{\ZZ}{\mathbb{Z}}$

• $L$ is not a split link. Equivalently, $X$ is irreducible. Equivalently, $\pi$ is not a free product.
• $L$ is not the unknot. Equivalently (as $L$ is not split) $\partial X$ is incompressible. Equivalently, $\pi$ is not $\ZZ$.
• $L$ has no cabled component, and has no parallel components. Equivalently, $X$ is acylinderical.
• $L$ has no component that is an "undisturbed satellite". Equivalently, $X$ is atoroidal. These last two topological properties are equivalent to $\pi$ not containing a copy of $\ZZ^2$.

Then $X$ admits a hyperbolic structure.

There is a topological criterion due to Thurston. Suppose the link $L$ is not split, is not the unknot, is not a torus knot, and is not a satellite knot. Then $L$ admits a hyperbolic structure as you describe. This is just part of a much bigger story, of course. Let me point you towards Wikipedia and the references therein.

http://en.wikipedia.org/wiki/Hyperbolic_link

There is a topological criterion due to Thurston. Using the JSJ machine (and work of many others) this criterion can also be phrased algebraically.

Suppose that $L$ is the link and $X$ is the link complement. Suppose $\pi = \pi_1(X)$. We assume the following. $\newcommand{\ZZ}{\mathbb{Z}}$

• $L$ is not a split link. Equivalently, $X$ is irreducible. Equivalently, $\pi$ is not a free product.
• $L$ is not the unknot. Equivalently (as $L$ is not split) $\partial X$ is incompressible. Equivalently, $\pi$ is not $\ZZ$.
• $L$ has no cabled component, and has no parallel components. Equivalently, $X$ is acylinderical.
• $L$ has no component that is an "undisturbed satellite". Equivalently, $X$ is atoroidal. These last two topological properties are equivalent to $\pi$ not containing a copy of $\ZZ^2$.

Then $X$ admits a hyperbolic structure as you describe. This is just part of a much bigger story, of course. Let me point you towards Wikipedia and the references therein.

http://en.wikipedia.org/wiki/Hyperbolic_link