I'm elementary in physics and I have a question about a notation. In the book, the author says that the rotation group is the set of linear transformations on $\mathbb{R}^n$ preserving the inner product $x^i x_i =\delta_{ij}x^i x^j$ and Lorentz transformations preserve the Minkowskian inner product $x^{\mu}x_{\mu}=g_{\mu \nu}x^{\mu}x^{\nu}$ where $x^i \to R_{ij}x_j$ and $x^i =\begin{pmatrix}x \\ y\end{pmatrix}$, and $x_i =(x\; y)$. How do we have $x^i x^j$ or $x^{\mu}x^{\nu}$? Aren't they column vectors? How does $x^i x_i$ convert to $\delta_{ij}x^i x^j$? and the last question that I have: What does the author mean by $$R_{ki}R_{lj}\delta_{kl}=\delta_{ij}?$$


2 Answers 2


How do we have $x^i x^j$ or $x^{\mu}x^{\nu}$? Aren't they column vectors?

$x^i$ is in itself a number, equal to one particular component of the column vector.

$x^i \hat{e}_i $ is the vector itself (number * vector = vector).

Colloquially we might just call $x^i$ a column vector but that's just loose language really, because in other circumstances we refer to it as a number.

Since $x^i$ is just a number we are free to multiply it by another number $x^j$. If $i=1,2,3$ then the result has 9 possibilities depending on the value of i,j.

How does $x^i x_i$ convert to $\delta_{ij}x^i x^j$?

This is called index raising or lowering. Multiplying a vector by the metric (a matrix) results in a quantity which transforms like the dual of that vector. So covariant turns into contravariant and vice versa. So we change the position of the index of the result:

$$\delta_{ij}x^i = x_j$$

The reason this works to give the inner product is really hand-in-hand with the definition of the metric. The metric was defined in such a way that if you do this and multiply the result by its contravariant counterpart, then you will get the inner product of the two vectors. While multiplying by other matrices could give you something that transforms as the dual, only the metric can give you the inner product like this.

and the last question that I have: What does the author mean by $$R_{ki}R_{lj}\delta_{kl}=\delta_{ij}?$$

Let's do the multiplication:

$R_{ki}R_{lj}\delta_{kl} = R_{li}R_{lj}$ (the delta makes $l=k$, in other words it multiplied by the identity matrix)

$=(R^T)_{il} R_{lj}$ (the transpose has inverted indices, now this has the form of regular matrix multiplication because the inner index is the same)

$=\delta_{ij}$ (the transpose of the rotation matrix is its inverse $R^T R = R R^T = I$. The components of the identity matrix are $\delta_{ij}$)

  • $\begingroup$ Thank you so much for your great explanation. I got it now properly. Appreciate that. $\endgroup$
    – M.Ramana
    Feb 28 at 6:21
  • $\begingroup$ Thanks :) I am glad it helped. Feel free to ask any followups if they come up. $\endgroup$ Feb 28 at 6:21
  • $\begingroup$ Sure. Thank you for your kindness :) $\endgroup$
    – M.Ramana
    Feb 28 at 6:45
  • $\begingroup$ Sorry. I have some questions. Inner product on $\mathbb{R}^3$ is a function $\mathbb{R}^3\times \mathbb{R}^3 \to \mathbb{R}$ with some conditions. How do we define the inner product of $x^i$ and $x^j$ as $x^i x_i =\delta_{ij}x^i x^j$? What does it mean? Similarly, how we define the Minkowski inner product of $x^{\mu}$ and $x^{\nu}$? $\endgroup$
    – M.Ramana
    Feb 28 at 8:21
  • $\begingroup$ The same question about the metric. Why $\delta_{ij}$ and $g_{\mu nu}$ are metris? The definition of a metric on $\mathbb{R}^4$ is a function $d:\mathbb{R}^4 \times \mathbb{R}^4 \to \mathbb{R}^{\geq 0}$ with some conditions. $\endgroup$
    – M.Ramana
    Feb 28 at 8:25

In tensor language, the metric, for example $\delta_{ij}$, is used to raise or lower indices when contracted with some tensor (in this case $x^{i}$ is a tensor of rank (1,0), i.e. a vector).

Therefore, when contracted the same indices (remember that repeated indices at different heights means an implicit sum), we obtain $$\delta_{ij}x^{i}x^j = x_jx^j$$

where i contracted the $i$'s.

Now if you want to generalize the idea of inner product to arbitrary dimensions $n$ you can build $g_{\mu\nu}x^\mu x^\nu$ , where $g_{\mu \nu}$ is the metric of your manifold and the greek indices run from $0$ to $n-1$ (the $0$-th coordinate is usually the timelike coordinate). In the special case of the Euclidean tridimensional space (the one of a lifetime), the metric is flat, i.e $\delta_{ij}$.

PD: This is my first answer, so i apologize if it is badly written.

  • $\begingroup$ Dear @10BlackHole , thank you so much for your answer and nice explanaion. I got it now properly.. Apprecaite it. $\endgroup$
    – M.Ramana
    Mar 2 at 5:32
  • $\begingroup$ I am glad you understood :). Happy I continue answering questions. $\endgroup$ Mar 2 at 5:38
  • $\begingroup$ That's your favor. Thanks so much again. $\endgroup$
    – M.Ramana
    Mar 2 at 5:42

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