# Differential geometry - strange formula

1. Feb 2, 2010

### player1_1_1

heeello friends!;]
i have book "wstęp do współczesnej geometrii różniczkowej" written by Konstanty Radziszewski, this title in english mean something like... "basics of modern differential geometry", and here are many formulas which look similar to one another, but I never know what it means and sometimes its difficult to understand this, examples:
$$A^{i^\prime}_iA^j_{i^\prime}=\delta^i_j$$
what this delta means? here is written that its 1 when i=j and 0 when $$i\neq j$$, but I dont know what it means, this multiplication value is just 1 or 0 depending on this value?
next problem:) I have formula for skalar multiplication of two vectors: $$\partial\mbox{x}(u)=\mbox{x}_i(u)\partial u_x^i,\quad\delta\mbox{x}(u)=\mbox{x}_i(u)\delta u_x^i$$
and the formula:
$$\partial\mbox{x}(u)\delta\mbox{x}(u)=\mbox{x}_i(u)\mbox{x}_j(u)\partial u_x^i\delta u_x^j=g_{ij}(u)\partial u_x^i\delta u_x^j$$
why in upper equation there is "i" in upper index, and then suddenly "j" in next formula? what happened after last equation? why there is suddenly $$g_{ij}(u)$$ instead of $$\mbox{x}_i(u)\mbox{x}_j(u)$$? what it means?

2. Feb 2, 2010

### Rasalhague

I'm relatively new to these ideas myself, and don't know what all of the notation means, but perhaps this answers part of your question. Hopefully someone more knowledgable than me can give you a better and more thorough explanation than I can.

You can think of $\delta^i_j$ (called "Kronecker's delta" or "the Kronecker delta") as the entries of the identity matrix, if you treat the upper index as the indicating the row, and the lower index as indicating the column: the nxn matrix with 1 for each entry of its main diagonal, and zeros everywhere else, where n is how many values an index can take. So for n = 3,

$$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$$

If i = j (row 1, column 1; row 2, column 2; row 3, column 3), then the entry is 1, otherwise it's zero (row 1, column 2, etc.).

For example, if i can be 1 or 2, then n = 2, and

$$A^i\enspace_k B^k\enspace_j = \delta^i_j$$

would stand for a set of n2 = 4 equations:

$$\sum_{k=1}^{2} A^1\enspace_k B^k\enspace_1 = \delta^1_1$$

$$\sum_{k=1}^{2} A^1\enspace_k B^k\enspace_2 = \delta^1_2$$

$$\sum_{k=1}^{2} A^2\enspace_k B^k\enspace_1 = \delta^2_1$$

$$\sum_{k=1}^{2} A^2\enspace_k B^k\enspace_2 = \delta^1_2$$

which we can express more neatly as a single matrix equation

$$AB = I = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$

where $I$ is the identity matrix.

$g_{ij}$ is called the metric tensor. If a and b are vectors, with components $a^i = \left( a^1, a^2, ..., a^n \right)$, and $b^j = \left( b^1,b^2,...,b^n \right)$, then their scalar product is

$$g_{ij}a^ib^j = a_jb^j = a^ib_i$$

which is just a quick way of writing

$$\sum_{i=1}^{n} \sum_{j=1}^{n} g_{ij}a^i b^j.$$