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Welcome back. You are rapidly approaching the actual subject of my lectures.

I wanted to stress, I mean I finished talking about how to express, to encode geometry into triangulations,

these white flags and spaces. And I want to emphasize again the key point of this vector formulation is that it works without ever introducing coordinates.

It doesn't need that and it's a very nice way of getting around some of the issues one has just in the standard continuum formulation

where one has to introduce coordinates in order to start writing down fields and start writing down the dynamics.

There was maybe, since I got quite a few questions afterwards, the point with introducing coordinates on a given triangle

didn't maybe get the right message I wanted to get across, completely across, which was the following point.

Of course I can, if I want to, I should say, why should I ever want to do this, reintroduce coordinates.

In general I will not be able to do it just on a large chunk of triangulation, but I will certainly be able to do it on a small neighborhood.

I will definitely be able to do it on an individual building block, so in two dimensions, on an individual triangle.

Whether I can extend a kind of standard coordinate patch to a larger neighborhood depends also on the curvature singularities I meet when I walk around in my triangulation.

So certainly, coordinate patches I introduce cannot have such singularities in their interior, then simply my map to an open set of Rn will break down,

so it will not be a good coordinate chart. But certainly, for instance, if you take two neighboring building blocks, say in 2D two neighboring triangles,

you can always even choose a common piece of R2, in this case open set, which covers both of them, and that's sometimes quite useful.

So the point I wanted to make about the individual triangle, I drew this in the following way.

So I said, look, the interior geometry is entirely given by specifying the three edge lengths. L1 squared, L2 squared, L3 squared.

Now imagine that I now wanted to reintroduce coordinates on this single triangle.

There of course is a zillion different ways of doing this. I picked just a specific way, you can argue that is for certain purposes, a nice way of doing it,

but it's totally arbitrary, I could do it in so many other ways.

So here I introduced a coordinate system which are called X and Y, which was not orthogonal obviously, because here I have not, that's not a right angle.

And the way this coordinate system will work is that here will be its origin, 0, 0, as I said before, here will be 1, 0, here will be the point 0, 1,

and this here, this will be lines of constant X, okay, constant X. And conversely, if I draw things parallel to the X axis, this will be lines of constant Y.

Okay, so constant Y. So each point here has now coordinate values X and Y, so I can characterize each single point.

Now I have introduced coordinates and I said already of course all the metric structure is specified.

Now with respect to these coordinates X and Y, the metric that this error invariant property, it's a flat matrix, that's how I defined it to be, will take a certain form,

G mu nu of X and Y. If I chose some other coordinate system, it will take a different explicit form.

Now the way I arranged things in this particular coordinate system was quite nice in the sense that the G mu nu I derived happened to not depend on X and Y,

but was a constant 2 by 2 matrix. And the entries I wrote down for you were like this, L1 squared, L2 squared in these corners, and then it had this 1 half L1 squared plus L2 squared minus L3 squared,

and the same down here, plus L2 squared minus L3 squared. Okay, so now why on earth would I wish to introduce, we introduce coordinates.

Well for example, imagine I've forgotten how to compute the volume of a triangle given its three edge lengths.

Okay, then what can we do? You know, we're theoreticians who say, all right, let me introduce coordinates, then there must be a G mu nu, like for instance the one I computed here,

and then let me compute the volume of the triangle as an integral over the triangle, dx dy square root of the determinant of the metric, and that will be the volume.

Okay, now I have a specific coordinate system given, right, so let me just do it. Now by definition here, my integration range for X and Y,

let's see, if I let it go from 0 to 1 both for X and Y, I get the area of a parallelogram, but it's of course exactly half this.

So let me put the factor one half here, 0 1, 0 1, and now, okay, now I should compute the determinant of this thing here,

so that will be, and takes the square root of it, so this will be L1 squared L2 squared minus one quarter L1 squared plus L2 squared minus L3 squared all squared, modulus square root.

Okay, so now you give me your triangle and then you'll just be able to do this computation. Of course, the result is totally independent of the choice of coordinate system I made, right,

so it just has that area as a function here of the Li squared.

Now let me even be more specific because later on when I talk about, when I start talking about causal dynamical triangulations, I will work with specific building blocks,

which have specific edge length assignments, and let me already kind of choose a coloring I would use later on also.

Also, in two dimensional CDT, I will work with triangles that have one space-like edge and two time-like edges, so the time-like edges are here in red,

the space-like edge has length squared Ls squared, and the time-like edges have a length squared Lt squared, and it's the same for both of these edges.

And the way I will arrange this thing is that all building blocks will also look identical, they will all be this building block, and they will just glue this building block.

Now, how, let me re-express, let me do something else. So, now we were saying before, what I will want to do in the end is I want to remove this regulator,

so I want to actually, I want to shrink the triangles away. So, let me identify, say, this Ls squared with some a squared, where I now think of a as some short-distance cutoff that in the end I want to send to zero.

So, this a plays the role of some short-distance UV cutoff. Now, this is now some other lengths, let me parameterize it in the following way,

let me write it as a minus alpha, alpha being now a positive real number, times Ls squared, so where alpha is larger than zero.

So, the minus sign, of course, takes care of the fact that, as I said before, these will be timelike edges, so they will have a negative length squared in the Minkowski matrix.

So, my, of course, my signature convention is minus plus, yeah, for minus for the time direction.