¿Cómo se demuestra que el laplaciano es el cuadrado del operador (euclidiano) de Dirac?

Question

Si entiendo correctamente, la distancia Euclídea Dirac operador está dado por

$$D=\sum_{i=1}^n e_i \frac{\partial}{\partial x_i},$$

donde $e_i$ son bases para $Cl_{0,n}(\mathbb{R})$, es decir, el $n$-dimensiones álgebra de Clifford negativo-definido firma a lo largo de los reales (por lo $e_i^2=-1$), y $x_i$ son las correspondientes coordenadas. Varias fuentes indican que $D^2 = -\Delta_n$ donde $\Delta_n$ es la norma Euclidiana operador de Laplace

$$\Delta_n = \sum_{i=1}^n \frac{\partial^2}{\partial x_i^2}.$$

Cuando yo escriba $D^2 f$ explícitamente para alguna función $f:\mathbb{R}^n \rightarrow \mathbb{R}$, escalar términos de la Laplaciano ciertamente aparecen, por ejemplo,

$$e_1 \frac{\partial}{\partial x_1}\left( e_1 \frac{\partial}{\partial x_1} f \right) = e_1 \left( e_1 \frac{\partial^2}{\partial x_1^2}f + \left(\frac{\partial}{\partial x_1}e_1\right)\frac{\partial}{\partial x_1}f \right)=e_1^2 \frac{\partial^2}{\partial x_1^2}f = -\frac{\partial^2}{\partial x_1^2}f.$$

Pero también me terminan con bivector cruz términos que no debería estar allí:

$$e_1 \frac{\partial}{\partial x_1}\left( e_2 \frac{\partial}{\partial x_2} f \right) = e_1 \left( e_2 \frac{\partial^2}{\partial x_1 \partial x_2}f + \left(\frac{\partial}{\partial x_1}e_2\right)\frac{\partial}{\partial x_2}f \right)=e_1 e_2 \frac{\partial^2}{\partial x_1 \partial x_2}f = e_{12}\frac{\partial^2}{\partial x_1 \partial x_2}f.$$

Debería ser sólo teniendo en cuenta los escalares parte de $D^2$, o estoy simplemente haciendo algo mal aquí?

Answer 1

Tenga en cuenta que

\begin{align*} D^2 &= \left(\sum_{i=1}^ne_i\frac{\partial}{\partial x_i}\right)^2\\ &= \left(\sum_{i=1}^ne_i\frac{\partial}{\partial x_i}\right)\left(\sum_{j=1}^ne_j\frac{\partial}{\partial x_j}\right)\\ &= \sum_{i=1}^ne_i\frac{\partial}{\partial x_i}\left(\sum_{j=1}^ne_j\frac{\partial}{\partial x_j}\right)\\ &= \sum_{i=1}^ne_i\sum_{j=1}^n\frac{\partial}{\partial x_i}\left(e_j\frac{\partial}{\partial x_j}\right)\\ &= \sum_{i=1}^ne_i\sum_{j=1}^ne_j\frac{\partial^2}{\partial x_i\partial x_j}\\ &= \sum_{i=1}^n\sum_{j=1}^ne_ie_j\frac{\partial^2}{\partial x_i\partial x_j}\\ &= \sum_{i=1}^n\left(\sum_{j<i}e_ie_j\frac{\partial^2}{\partial x_i\partial x_j} + e_i^2\frac{\partial^2}{\partial x_i^2} + \sum_{j>i}e_ie_j\frac{\partial^2}{\partial x_i\partial x_j}\right)\\ &= \sum_{i=1}^n\sum_{j<i}e_ie_j\frac{\partial^2}{\partial x_i\partial x_j} + \sum_{i=1}^ne_i^2\frac{\partial^2}{\partial x_i^2} + \sum_{i=1}^n\sum_{j>i}e_ie_j\frac{\partial^2}{\partial x_i\partial x_j}\\ &= \sum_{i=1}^n\sum_{j<i}e_ie_j\frac{\partial^2}{\partial x_i\partial x_j} + \sum_{i=1}^n(-1)\frac{\partial^2}{\partial x_i^2} + \sum_{j=1}^n\sum_{i<j}e_ie_j\frac{\partial^2}{\partial x_i\partial x_j}\\ &= \sum_{i=1}^n\sum_{j<i}e_ie_j\frac{\partial^2}{\partial x_i\partial x_j} - \sum_{i=1}^n\frac{\partial^2}{\partial x_i^2} + \sum_{i=1}^n\sum_{j<i}e_je_i\frac{\partial^2}{\partial x_j\partial x_i}\\ &= \sum_{i=1}^n\sum_{j<i}e_ie_j\frac{\partial^2}{\partial x_i\partial x_j} - \Delta_n + \sum_{i=1}^n\sum_{j<i}e_je_i\frac{\partial^2}{\partial x_i\partial x_j}\\ &= \sum_{i=1}^n\sum_{j<i}e_ie_j\frac{\partial^2}{\partial x_i\partial x_j} - \Delta_n + \sum_{i=1}^n\sum_{j<i}(-e_ie_j)\frac{\partial^2}{\partial x_i\partial x_j}\\ &= \sum_{i=1}^n\sum_{j<i}e_ie_j\frac{\partial^2}{\partial x_i\partial x_j} - \Delta_n - \sum_{i=1}^n\sum_{j<i}e_ie_j\frac{\partial^2}{\partial x_i\partial x_j}\\ &= -\Delta_n. \end{align*}