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@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, $$ D=\sigma _x\partial_x + \sigma_y \partial_y = \begin{pmatrix} 0&\partial_x -i\partial_y\\ \partial_x+i\partial_y&0 \end{pmatrix} ~~~\leadsto \\ D^2 = \Delta_2 ~~{\mathbb 1}_2. $$ 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, $$ D=\sigma _x\partial_x + \sigma_y \partial_y + \sigma_z \partial_z = \begin{pmatrix} \partial_z&\partial_x -i\partial_y\\ \partial_x+i\partial_y&-\partial_z \end{pmatrix} ~~~\leadsto \\ D^2 = \Delta_3 ~~{\mathbb 1}_3. $$

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.

• 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.

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

@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, $$ D=\sigma _x\partial_x + \sigma_y \partial_y = \begin{pmatrix} 0&\partial_x -i\partial_y\\ \partial_x+i\partial_y&0 \end{pmatrix} ~~~\leadsto \\ D^2 = \Delta_2 ~~{\mathbb 1}_2. $$ 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, $$ D=\sigma _x\partial_x + \sigma_y \partial_y + \sigma_z \partial_z = \begin{pmatrix} \partial_z&\partial_x -i\partial_y\\ \partial_x+i\partial_y&-\partial_z \end{pmatrix} ~~~\leadsto \\ D^2 = \Delta_3 ~~{\mathbb 1}_3. $$

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, $D= -\vec \partial \cdot ~ (\sigma_2\otimes\vec \sigma )+ \partial_w ~\sigma_1\otimes {\mathbb 1}_2$, 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...)

@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, $$ D=\sigma _x\partial_x + \sigma_y \partial_y = \begin{pmatrix} 0&\partial_x -i\partial_y\\ \partial_x+i\partial_y&0 \end{pmatrix} ~~~\leadsto \\ D^2 = \Delta ~~{\mathbb 1}_2. $$ 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, $$ D=\sigma _x\partial_x + \sigma_y \partial_y + \sigma_z \partial_z = \begin{pmatrix} \partial_z&\partial_x -i\partial_y\\ \partial_x+i\partial_y&-\partial_z \end{pmatrix} ~~~\leadsto \\ D^2 = \Delta ~~{\mathbb 1}_3. $$

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

@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, $$ D=\sigma _x\partial_x + \sigma_y \partial_y = \begin{pmatrix} 0&\partial_x -i\partial_y\\ \partial_x+i\partial_y&0 \end{pmatrix} ~~~\leadsto \\ D^2 = \Delta_2 ~~{\mathbb 1}_2. $$ 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, $$ D=\sigma _x\partial_x + \sigma_y \partial_y + \sigma_z \partial_z = \begin{pmatrix} \partial_z&\partial_x -i\partial_y\\ \partial_x+i\partial_y&-\partial_z \end{pmatrix} ~~~\leadsto \\ D^2 = \Delta_3 ~~{\mathbb 1}_3. $$

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.

• 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.

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

@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, $$ D=\sigma _x\partial_x + \sigma_y \partial_y = \begin{pmatrix} 0&\partial_x -i\partial_y\\ \partial_x+i\partial_y&0 \end{pmatrix} ~~~\leadsto \\ D^2 = \Delta ~~{\mathbb 1}_2. $$ 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, $$ D=\sigma _x\partial_x + \sigma_y \partial_y + \sigma_z \partial_z = \begin{pmatrix} \partial_z&\partial_x -i\partial_y\\ \partial_x+i\partial_y&-\partial_z \end{pmatrix} ~~~\leadsto \\ D^2 = \Delta ~~{\mathbb 1}_3. $$

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 space like ones with i, and dot them with the d-dimensional gradient to factorize your Laplacian in all dimensions, as above. Again, I'm not clear what you imagine you can get for CFTs in all dimensions this way. (To me the gig looks useless, but I don't wish to discourage creative thinking...)

@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, $$ D=\sigma _x\partial_x + \sigma_y \partial_y = \begin{pmatrix} 0&\partial_x -i\partial_y\\ \partial_x+i\partial_y&0 \end{pmatrix} ~~~\leadsto \\ D^2 = \Delta ~~{\mathbb 1}_2. $$ 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, $$ D=\sigma _x\partial_x + \sigma_y \partial_y + \sigma_z \partial_z = \begin{pmatrix} \partial_z&\partial_x -i\partial_y\\ \partial_x+i\partial_y&-\partial_z \end{pmatrix} ~~~\leadsto \\ D^2 = \Delta ~~{\mathbb 1}_3. $$

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