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[Martin Sandve Alnæs <martinal@xxxxxxxxx>]

It's working, from the test:

Note that e here will be a 3x3x3 tensor which must be indexed to get eps_ijk
        e = PermutationSymbol(3)
        self.assertEqual(e.shape(), (3, 3, 3))

Evaluation with int indices produce correct values for eps_ijk:
        for i in range(3):
            for j in range(3):
                for k in range(3):
                    value = (j-i)*(k-i)*(k-j)/2
                    self.assertEqual(e[i, j, k], value)

If e is indexed with free indices, we get an Indexed object as usual in ufl
        i, j, k = indices(3)
        self.assertIsInstance(e[i,j,k], Indexed)

and evaluating the product eps_ijk eps_ijk with implicit summation gives the
right sum
        x = (0,0,0)
        self.assertEqual((e[i,j,k] * e[i,j,k])(x), 6)

I haven't tested with n != 3, but the formula Garth copied should handle
that.

Martin


On 14 April 2011 09:19, Mikael Mortensen <mikael.mortensen@xxxxxxxxx> wrote:

> The source of truth suggests Levi-Civita, alternating or permutation
> symbol: http://en.wikipedia.org/wiki/Levi-Civita_symbol
>
> Mikael
>
>
> On 13 April 2011 20:31, Martin Sandve Alnæs <martinal@xxxxxxxxx> wrote:
>
>> That should be fairly easy to do, can be handled similarly to Identity.
>>
>> I = Identity(cell.d)
>> f = I[i,j] * ...
>>
>> e = Permutation(cell.d)
>> f = e[i,j,l] * v[j]*w[i]*v[l]*dx
>>
>> Other suggestions for naming?
>>
>> Martin
>>
>> On 13 April 2011 20:24, Garth N. Wells <gnw20@xxxxxxxxx> wrote:
>>
>>> The permutation symbol should be added to UFL.
>>>
>>> Garth
>>>
>>> On 13/04/11 19:20, Martin Sandve Alnæs wrote:
>>> > Python does not know anything about implicit index summation, so in
>>> your
>>> > code e() is only called once, and reaches the else condition and thus
>>> > returns 0.0.
>>> >
>>> > Similar to as_vector, you can use the as_tensor function, since you
>>> need
>>> > three dimensions.
>>> >
>>> > Martin
>>> >
>>> > On 13 April 2011 19:04, B. Emek Abali
>>> > <question152702@xxxxxxxxxxxxxxxxxxxxx
>>> > <mailto:question152702@xxxxxxxxxxxxxxxxxxxxx>> wrote:
>>> >
>>> >     New question #152702 on DOLFIN:
>>> >     https://answers.launchpad.net/dolfin/+question/152702
>>> >
>>> >     May be a non-needed trial for some, but I try to define an
>>> >     expression (permutation symbol)
>>> >
>>> >     def e(i,j,k):
>>> >            if (i,j,k) == (1.0,2.0,3.0): return 1.0
>>> >            elif (i,j,k) == (2.0,3.0,1.0): return 1.0
>>> >            elif (i,j,k) == (3.0,1.0,2.0): return 1.0
>>> >            elif (i,j,k) == (3.0,2.0,1.0): return -1.0
>>> >            elif (i,j,k) == (2.0,1.0,3.0): return -1.0
>>> >            elif (i,j,k) == (1.0,3.0,2.0): return -1.0
>>> >            else: return 0.0
>>> >
>>> >     and use (with summation over indices, it is a curl operator
>>> >     contracted with w) in the form like
>>> >
>>> >     f=v[j]*w[i]*e(i,j,l)*v[l]*dx
>>> >
>>> >     which raises:
>>> >     ufl.log.UFLException: Trying to integrate expression of rank 0 with
>>> >     free indices
>>> >
>>> >     Do I have to use as_vector(...) type definition to sum with indices
>>> >     or is there a nice possibility to let it sum over the arguments of
>>> >     e(...) expression?
>>> >
>>> >     --
>>> >     You received this question notification because you are a member of
>>> >     DOLFIN Team, which is an answer contact for DOLFIN.
>>> >
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>>
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>