So now we have to start from the beginning.

The Stokes theorem allows you to switch from an interval along a circuit to an interval over the enclosed area.

So the knot is the area that is enclosed in the circuit before the deformation.

The dA is the vector-valued area element.

The operator is the gradient of the divergence.

The rotation is essentially the derivative of the component that we look at.

So in this case, the rotation of F or the curl of the component is taking the gradient of F, so F is the second order tensor, so the gradient is then a third order tensor, and then protecting the symmetry of the last two units.

And that is done by this operation with double quantum with the third order commutation symbol.

We have the same letter here as for the constant, so don't execute that.

Specifically, the third order commutation symbol that has 1 and 1.0.

And then if we spell that out in indices, we have Fij, the gradient of F, comma k, and then this double dot is the double construction of the gradient.

At the end of the day, we get a tensor that has indices with little l in the l, which is the gradient of the tensor field.

The order of the tensor is lower than the order of the tensor, and the curl of the page is the length of the second order tensor.

So we have the second order tensor, two-point tensor, and the curl of the page is also the second order tensor in this case, or two-point tensor in the lower case.

And then we try to see whether we have our definitions of this whole theory or not, as a little bit of a matter of days, different definitions in the set of at the end of the day as long as we have different definitions.

Okay, now, if you try to understand this better, you see the jk is double construction with this jk.

So we have two of the properties of this E jkl, so that is 1, and jkl is 1, 2, 3, 3, 3, 1, 3, 1, 2, and this is one of the connotations of 3, 2, 1, and minus 1, and in all other cases, 0.

So at the end of the day, what you can get here is that essentially with double construction, the symmetry in this line is too infinite.

So if you try to write it in a different way, if we take this two-symmetric part of this equation, so that will be jk minus kj, that will be two-symmetric contributions, and then we will have the function 1.

And this is typically abbreviated by square matrices.

So we abbreviate symmetric contributions or symmetric combinations of two indices by a round vector.

So what this line tells us is that whether you put f i j comma 10 or its two symmetric contributions, that is the same thing.

So we have many symmetric contributions to this index section here, which means filters out the other constructions which are connotations in this.

So if there is a double construction with this E jk, anything that is symmetric in the jk, that will constitute 0, because the function is a block one, and the function that comes with the number one, is simply cancelled.

So at the end of the day, we check whether there is any non-symmetry content here in the last two integers.

So once again, we start with asking ourselves if there is a remaining order failure, if I do the circle integral, I invoke the relations between the line elements, ask them for the deformations, and I invoke both theorems.

And then again, since any surface that we can think of as a soft-turn within this entire loop here, is satisfied with the relation, plus we are requiring that for any other surface, we don't want to have to use any snow there,

this requirement that the integral is 0 can be localized, and we ask for the quantity here to integrate with the curl of M, and this benefits us.

Okay? Okay, and then if we analyze the meaning of the curl of the rotation, we see that essentially the supermetric part of this graph should be independent.

Alright, let's consider a case where x is a loop, so there it derives from a corresponding deformation f, so f by j is computed from y i comma j, okay?

If you put another gradient on the f here, you get this f by j comma k, from this expression here, expressed in front of the deformation f, this y i, then the technical derivative j k.

The technical derivative of the technical derivative are per se symmetric, so you can explain this with sequence. Whether you first take the derivative of x k, then you can take the x k of the observer that gives the

different things. So in the case of a compositive deformation gradient, f by j comma k, it also gives the same expression as f by j comma k, which continues with the order of the

derivative, which is to say that f by j k is symmetric in j k, or the supermetric part is zero. And this is exactly what this shows in the case of the AEPS, whether or not the

supermetric is zero. So if you don't have that, then we can deduce that we can derive f in each of the deformations that we find. So point out locally, we can deduce from this that the

condition that f is indeed because it's compatible, that it fits into a certain continuous deformation, that at every point in our body, this quantity is the rotation of f in each of the deformations.

Okay. Okay, there are some particularities here involved that have to do with the topology of the body, particularly for the supermetric connection body, and the conditions that are

involved in the supermetric. So since this is such an important quantity, we have given it a separate name, and called it A. And A is just a placeholder for a quantity that is

simply stepping off the crystal line material that we noted in so-called locations and locations of the density kind of. So if you look at this picture here, you can see what a crystal line material looks like

in this case. And then you see here all the atoms at the end of this circle. You see the atoms here, combining with the atoms. Okay. And then you see here a typical dislocation,

and dislocation, which is defined as the following node of your exosynaptic line, defined by one of the lattice lines as an N-beam. Okay. So then if you take the corresponding thing to this

chicken, take it to the bottom of this circle here, and then you see the

Then we go three steps down.

One, two, three.

The start of here we end up here.

So that's essentially in a crystal line.

That is the dual position that I mean.

Now, you go around the globe.

And whatever, let's do it again.

We start here.