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So I have already know that the fundamental group of figure-eight knot is given as: $G=(a,d\mid ada^{-1}da=dad^{-1}ad)$, where a,d are the two generators, then I have searched many books, but didn't find one explicit homomorphism $\rho$ that sends $G$ to $SL_{2}(\mathbb{C})$, the only thing they mentioned is that this map should send

$\rho$(a)=$\left(\begin{array}{rr} 1&1 \\ 0&1 \\ \end{array}\right)$ and $\rho$(d)=$\left(\begin{array}{rr} 1&0 \\ -\omega&1 \\ \end{array}\right)$

So I am wondering what will the explicit homomorphism be? Does this has anything to do with the character of the representation?

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 How much more explicit do you want it to be? Any element of $G$ can be expressed as a word in $a$ and $d$. Pick such an expression, and replace each occurrence of $a,d$ with the two matrices, and perform the matrix multiplication. You may want to recall that one property of finite presentations of groups is that it is enough to give images on generators to get a fully described homomorphism. – Mikael Vejdemo-Johansson May 15 2011 at 9:00
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 Presumably the question is "what is $\omega$?". The answer, in detail, is in Riley's article qjmath.oxfordjournals.org/content/35/2/…. – Paul May 15 2011 at 13:34
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 You write $\rho(b)$. Did you mean $\rho(d)$? – David Speyer May 15 2011 at 15:30
$\omega$ is usually the primitive cube root of unity. – Igor Rivin May 15 2011 at 17:34

closed as too localized by Martin Brandenburg, José Figueroa-O'Farrill, Ryan Budney, Zev Chonoles, Daniel Litt May 15 2011 at 18:12

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