Factoring the Laplace operator $Delta$ in dimensions $D geq 3$

Question

Consider the Laplace operator in 2 dimensions
begin{equation}
Delta = frac{partial^2}{partial  x^2} + frac{partial^2}{partial  y^2} = partial^2_x + partial^2_y
end{equation}
By defining the complex variables $z$ and $bar{z}$
begin{equation}
begin{split}
z &= x+iy
bar{z} &= x-iy
end{split}
end{equation}
we can "factor" the Laplace operator into the product of 2 first order differential operators using freshman calculus
begin{equation}
Delta = 4 partial_z  partial_{bar{z}}
end{equation}
The other day I was watching the lecture Part 1 | Introduction to conformal field theory: Liouville model | Leon Takhtajan | Лекториум, when the lecturer remarked that this "factoring" property of the Laplace operator in 2 dimensions lies in the heart of Conformal Field Theory and Complex Analysis. He then said that this factoring bussiness can also be realized in dimensions $D geq 3$, albeit in a much more involved manner, using spinors and Dirac matrices.
My question is, how can we factor the Laplace operator in dimensions $D geq 3$, using spinors and Dirac matrices? What does this tell us about Conformal Field Theories in $D geq 3$?

Cosmas Zachos · Answer

@AccidentalFourierTransform drew your attention to the Dirac operator, which, of course, can go Euclidean and extend to all dimensions. It has little to do with conformal FT, if that's where you want to go, because the conformal group is finite in all dimensions different than d=2. So disaggregate the conformal gig.
First appreciate that Leon's construction can be superfluously replicated to 2×2 matrices (Pauli, duh!), through defining the Euclidean 2d Dirac operator,
$$
D=sigma _xpartial_x + sigma_y partial_y =    begin{pmatrix}
      0&partial_x -ipartial_y
      partial_x+ipartial_y&0
    end{pmatrix} ~~~leadsto 
D^2 = Delta_2 ~~{mathbb 1}_2. 
$$
So far superfluous, but you may extend this to 3d, again with 2×2 matrices since their spinors are the same for an even dimension and its odd-one-higher one,
$$
D=sigma _xpartial_x + sigma_y partial_y + sigma_z partial_z =    begin{pmatrix}
      partial_z&partial_x -ipartial_y
      partial_x+ipartial_y&-partial_z
    end{pmatrix} ~~~leadsto 
D^2 = Delta_3 ~~{mathbb 1}_3.
$$
And so on. The link provided will let you write down the Dirac gamma matrices in all dimensions, rectify your Minkowski metric by multiplying the spacelike ones with i, and dot them with the d-dimensional gradient to factorize your Laplacian in all dimensions, as above. For instance, for d =4, $D= -vec partial cdot ~ (sigma_2otimesvec sigma )+ partial_w ~sigma_1otimes {mathbb 1}_2$, hence, yet again, $D^2 = Delta_4 ~~{mathbb 1}_4$.

Leon's "much more involved" refers to the tensor product structure of coordinate space with spinor space. This latter part you cannot eschew, and it is a misunderstanding to expect it to be absent from the picture.

I'm not clear what you imagine you could get for CFTs in all dimensions this way. (To me the gig looks useless, but I should not aim to discourage creative thinking...)

Factoring the Laplace operator $Delta$ in dimensions $D geq 3$

One Answer

Add your own answers!

Ask a Question