Ngô Quốc Anh

January 26, 2010

RG: Pointwise norm of an (r,s)-tensor relative to a metric g

Filed under: Nghiên Cứu Khoa Học, Riemannian geometry — Tags: — Ngô Quốc Anh @ 1:03

Today, let’s talk about the pointwise norm of an $(r,s)$-tensor $\alpha$ relative to a metric $g$. We assume $\displaystyle\alpha= \alpha _{{j_1} \cdots {j_r}}^{{k_1} \cdots {k_s}}\frac{\partial }{{\partial {x^{{k_1}}}}} \otimes\cdots\otimes \frac{\partial }{{\partial {x^{{k_s}}}}} \otimes d{x^{{j_1}}} \otimes\cdots\otimes d{x^{{j_r}}}$.

Then $\displaystyle \left| \alpha\right| = \sqrt {{g^{{j_1}{q_1}}}...{g^{{j_r}{q_r}}}{g_{{k_1}{p_1}}}...{g_{{k_s}{p_s}}}\alpha _{{j_1} \cdots {j_r}}^{{k_1} \cdots {k_s}}\alpha _{{q_1} \cdots {q_r}}^{{p_1} \cdots {p_s}}}$.

For example, if $f$ is a scalar function, then $\nabla f$ is an $(1,0)$-tensor and $\nabla^2 f$ is an $(2,0)$-tensor. To be precise, $\displaystyle \nabla f = \frac{{\partial f}}{{\partial {x^i}}}d{x^i}, \quad {\nabla ^2}f = \left( {\frac{{{\partial ^2}f}}{{\partial {x^i}\partial {x^j}}} - \Gamma _{ij}^k\frac{{\partial f}}{{\partial {x^k}}}} \right)d{x^i} \otimes d{x^j}$.

Then $\displaystyle\left| {\nabla f} \right| = \sqrt {{g^{ij}}\frac{{\partial f}}{{\partial {x^i}}}\frac{{\partial f}}{{\partial {x^j}}}}$

and $\displaystyle\left| {{\nabla ^2}f} \right| = \sqrt {{g^{ip}}{g^{jq}}\left( {\frac{{{\partial ^2}f}}{{\partial {x^i}\partial {x^j}}} - \Gamma _{ij}^k\frac{{\partial f}}{{\partial {x^k}}}} \right)\left( {\frac{{{\partial ^2}f}}{{\partial {x^p}\partial {x^q}}} - \Gamma _{pq}^l\frac{{\partial f}}{{\partial {x^l}}}} \right)}$.