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Can anybody explain why Levi-Civita Christoffel symbol in general $N$ dimensional space have $\frac{N^2(N+1)}{2}$ independent components? I have read that in $N$-dimensional space, metric tensor has at most $\frac{N(N+1)}{2}$ independent components and for each component we can have $N$ independent christoffels and thus one arrives at the final expression. But I can't understand how these numbers are arising. Kindly elaborate or suggest any book where I can get the detalied discussion on this.

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First, let's talk about the metric tensor; it has $\frac{N(N+1)}{2}$ independent components, because it is a symmetric tensor: $g_{ab}=g_{ba}$. Writing this out in a matrix format: \begin{align} [g]&= \begin{pmatrix} g_{11}&\dots& g_{1N}\\ \vdots &\ddots&\vdots\\ g_{N1}&\dots & g_{NN}. \end{pmatrix} \end{align} In this format, everything which lies on the main diagonal and the upper triangle are the independent components, because those in the strictly lower triangle are determined by symmetry. So, you just count how many such guys there are; it is easily seen to be $\frac{N(N+1)}{2}$. For example, $N$ independent entries in the first row, $N-1$ in second row, ..., $1$ in the last row. An alternate way of counting them is that there are a total of $N^2$ entries; if we ignore the $N$ of them which are on the main diagonal, that leaves us with $N^2-N$ components; but by symmetry only half are independent. So, the total number of independent components is $N+\frac{N^2-N}{2}=\frac{N(N+1)}{2}$ (which is just another way of saying that $1+2+\dots +N=\frac{N(N+1)}{2}$).

Next, the Christoffel symbols $\Gamma^{a}_{bc}$ have $3$ indices. Each index $a,b,c$ can take on any value between $1$ and $N$. So, a-priori, that's a total of $N^3$ functions. However, if you consider torsion-free connections (which is the case for the Levi-Civita connection, which comes from the metric), then we have symmetry in the lower entries, so $\Gamma^a_{bc}=\Gamma_{cb}$. That reduces the number of components down to $N\cdot\frac{N(N+1)}{2}=\frac{N^2(N+1)}{2}$ components ($N$ choices for $a$, and $\frac{N(N+1)}{2}$ for the lower slots $b,c$, taking symmetry into account).

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  • $\begingroup$ Thanks a lot! Why is there $\frac{N(N+1)}{2} $ choices for the lower indices? $\endgroup$
    – Math noob
    Jun 30 at 14:53
  • $\begingroup$ @Mathnoob I feel like I explained that in the first paragraph $\endgroup$
    – peek-a-boo
    Jun 30 at 14:55
  • $\begingroup$ I mean how does the number of independent components of metric affect the choice of $b$ and $c$? $ \Gamma^{a}_{bc} = \frac{1}{2} g^{a \mu}[...]$. Now when $a$ is fixed, we can choose $\mu$ in $\frac{N(N+1)}{2}$ ways as there only $\frac{N(N+1)}{2}$ independent $g^{a \mu}$. Am I correct? $\endgroup$
    – Math noob
    Jun 30 at 15:04
  • $\begingroup$ @Mathnoob forget about the explicit formula for $\Gamma$ in terms of $g$. $\Gamma^{a}_{bc}$ is just an object with $3$ indices. Fix $a=1$ for example. Then $\Gamma^{1}_{bc}$ is an object with $2$ indices, and it happens to be symmetric (for torsion-free connections). My first paragraph's counting arguemnt applies to any doubly-indexed symmetric object; not just the metric tensor. Thus, $\Gamma^1_{bc}$ has $\frac{N(N+1)}{2}$ independent choices. Finally, repeat for $a\in\{1,\dots, N\}$ to get $N\cdot\frac{N(N+1)}{2}$ independent choices for $\Gamma^a_{bc}$. $\endgroup$
    – peek-a-boo
    Jun 30 at 15:26
  • $\begingroup$ Yup, got it! Tons of thanks again @peek-a-boo $\endgroup$
    – Math noob
    Jun 30 at 15:47

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