SciForums.com > Science > Physics & Math > Quaternions PDA View Full Version : Quaternions Post ReplyCreate New Thread arfa brane05-12-12, 06:17 PMTo show H = {1,-1,i,-i,j,-j,k,-k} forms a group under multiplication, need to show that the operation is associative. Then show there is an identity (which should be easy seeing as how 1 is an element of H), and that every nonzero element (quaternion) has an inverse. The associative part is also easy, apparently. You just use a basis which is the matrix form of H, since matrix multiplication is associative, and we're done? But how do you find (other than looking it up) the matrix representation of each element in H? 1\; =\; \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} is straightforward, so -1 is too. What about the rest? 1100f05-12-12, 08:32 PMTo show H = {1,-1,i,-i,j,-j,k,-k} forms a group under multiplication, need to show that the operation is associative. Then show there is an identity (which should be easy seeing as how 1 is an element of H), and that every nonzero element (quaternion) has an inverse. The associative part is also easy, apparently. You just use a basis which is the matrix form of H, since matrix multiplication is associative, and we're done? But how do you find (other than looking it up) the matrix representation of each element in H? 1\; =\; \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} is straightforward, so -1 is too. What about the rest? Pauli matrices arfa brane05-12-12, 10:42 PMActually I realised that you only need to show that quaternion multiplication forms a distributive set of linear terms, so each term depends linearly on the coefficients of each quaternion (this is in the lecture notes). So there is a matrix representation, and hence multiplication is associative. So you don't need to explicitly describe the matrices, just show they must exist. In fact, this is well known. That there are four linear terms for each component of the product implies a 4x4 matrix representation. AlphaNumeric05-13-12, 03:48 AMObviously there's more elegant ways to do it but it is possible to do it brute force from the axioms, if you don't know any applicable results from other areas of maths. The axioms are ij=k, jk=i, ki=j, ii=jj=kk=-1, ji=-k , kj=-i , ik=-j. Now you could just write down every possible triplet a,b,c and show by hand (ab)c = a(bc) but there's a more compact form which doesn't require anything beyond the axioms. If we call \sigma_{1},\sigma_{2},\sigma_{3} = i,j,k then \sigma_{a}\sigma_{b} = 1\delta_{ab} + \epsilon_{abc}\sigma_{c}[/tex. Of course it might not be obvious that they can be written like that, I'm cheating a little because I know the 4 matrices form a representation of the sl(2,C) Lie algebra basis but like I said, you can brute force it but I'm going to 'pseudo-brute force' it by using this expression to represent multiple cases. If you combine the brute force method with the cyclic permutation symmetry of the axioms you'll effectively reduce the problem to this anyway. Now we want to show [tex]\sigma_{a}(\sigma_{b}\sigma_{c}) = (\sigma_{a}\sigma_{b})\sigma_{c}. Well now we can write the pair in each bracket as a linear combination of the basis elements, LHS : \sigma_{a}(\sigma_{b}\sigma_{c}) = \sigma_{a}(1\delta_{bc} + \epsilon_{bcd}\sigma_{d}) = \sigma_{a}\delta_{bc} + \epsilon_{bcd}(\sigma_{a}\sigma_{d}) = \sigma_{a}\delta_{bc} + \epsilon_{bcd}(1\delta_{ad} + \epsilon_{ade}\sigma_{e}) = \epsilon_{bca} + \sigma_{a}\delta_{bc} + \epsilon_{bcd}\epsilon_{ade}\sigma_{e} = \epsilon_{abc} + \sigma_{a}\delta_{bc} + (\delta_{be}\delta_{ca} - \delta_{ba}\delta_{ce})\sigma_{e} = \epsilon_{abc} + \sigma_{a}\delta_{bc} + \delta_{ca}\sigma_{b} - \delta_{ba}\sigma_{c} RHS : (\sigma_{a}\sigma_{b})\sigma_{c} = (\delta_{ab} + \epsilon_{abd}\sigma_{d})\sigma_{c} = \delta_{ab}\sigma_{c} + \epsilon_{abd}(\delta_{dc} + \epsilon_{dce}\sigma_{e}) = \delta_{ab}\sigma_{c} + \epsilon_{abc} + (\delta_{ac}\delta_{be} - \delta_{ae}\delta_{bc})\sigma_{e} = \delta_{ab}\sigma_{c} + \epsilon_{abc} + \delta_{ac}\sigma_{b} - \delta_{bc}\sigma_{a} Since \delta_{ab} = \delta_{ba} we're done :) Obviously even with this method it's still not terribly pleasant but that's usually what happens if you don't know or can't use some more general result. Post ReplyCreate New Thread