How to construct the charge conjugation matrix for any given spacetime dimension? Generally, Gamma matrices could be constructed based on the Clifford algebra.
\begin{equation}
\gamma^{i}\gamma^{j}+\gamma^{j}\gamma^{i}=2h^{ij},
\end{equation}
My question is how to generally construct the charge conjugation matrix to raise one spinor index in the gamma matrix.
In even dimensions (D=2m), consider
complex Grassmann algebra $\Lambda_{m}[\alpha^{1},...,\alpha^{m}]$ with
generators $\alpha^{1},...,\alpha^{m}.$) Namely, we define $\widehat{\alpha
}^{i}$ and $\widehat{\beta}_{i}$ as multiplication and differentiation
operators:
\begin{equation}
\widehat{\alpha}^{i}\psi=\alpha^{i}\psi,
\end{equation}
\begin{equation}
\widehat{\beta}_{i}\psi=\frac{\partial}{\partial\alpha^{i}}\psi.
\end{equation}
According to the Grassmann algebra, we have
\begin{equation}
\widehat{\alpha}^{i}\widehat{\alpha}^{j}+\widehat{\alpha}^{j}\widehat{\alpha
}^{i}=0,
\end{equation}
\begin{equation}
\widehat{\beta}_{i}\widehat{\beta}_{j}+\widehat{\beta}_{j}\widehat{\beta}%
_{i}=0
\end{equation}
\begin{equation}
\widehat{\alpha}^{i}\widehat{\beta}_{j}+\widehat{\beta}_{j}\widehat{\alpha
}^{i}=\delta_{j}^{i}.
\end{equation}
This means that $\widehat{\alpha}^{1},...,\widehat{\alpha}^{m},\widehat{\beta
}_{1},...,\widehat{\beta}_{m}$ specify a representation of Clifford algebra
for some choice of $h$ (namely, for $h$ corresponding to quadratic form
$\frac{1}{2}(x^{1}x^{m+1}+x^{2}x^{m+2}+...+x^{m}x^{2m})$). It follows that
operators
\begin{equation}
\Gamma^{j}=\widehat{\alpha}^{j}+\widehat{\beta}_{j},1\leq j\leq m,
\end{equation}
\begin{equation}
\Gamma^{j}=\widehat{\alpha}^{j-m}-\widehat{\beta}_{j-m},m<j\leq2m,
\end{equation}
determine a representation of $Cl(m,m,\mathbb{C})$.
For example, in $D=4$, we can obtain
      $$\Gamma^{1}=\begin{pmatrix}0&
   1&
   0&
   0\\
   1&
   0&
   0&
   0\\
   0&
   0&
   0&
   1\\
   0&
   0&
   1&
   0\\
   \end{pmatrix}$$,
      $$\Gamma^{2}=\begin{pmatrix}0&
   0&
   0&
   1\\
   0&
   0&
   {-1}&
   0\\
   0&
   {-1}&
   0&
   0\\
   1&
   0&
   0&
   0\\
   \end{pmatrix}$$,
$$\Gamma^{3}=\begin{pmatrix}0&
   {-1}&
   0&
   0\\
   1&
   0&
   0&
   0\\
   0&
   0&
   0&
   1\\
   0&
   0&
   {-1}&
   0\\
   \end{pmatrix}$$,
$$\Gamma^{4}=\begin{pmatrix}0&
   0&
   0&
   {-1}\\
   0&
   0&
   1&
   0\\
   0&
   {-1}&
   0&
   0\\
   1&
   0&
   0&
   0\\
   \end{pmatrix}.$$
My question is how to generally construct the charge conjugation matrix C, so that we could have
$$C\Gamma C^{-1}=\pm\Gamma^T$$
 A: Explicit expressions for the Euclidian signature are given in the following Hitoshi Murayama lecture notes  (Section 1.3). The expressions are given in the Pauli matrix tensor product basis.
A: The charge conjugation matrix will depend on the choice of the basis you are representing the Dirac matrix. This is so because you want to satisfy:
$$
C\Gamma^{m}C^{-1}=\pm(\Gamma^{m})^{T}
$$
Two charge conjugation matrices of different choices of basis will be related by 
$$
C\rightarrow U^{T}CU
$$
where the Dirac matrices of different basis will be related by 
$$
\Gamma^{m}\rightarrow U\Gamma^{m} U^{-1}
$$
Now, you need to fix a basis and find the charge conjugation matrix for this basis. 
There is a very convenient basis obtained by splitting the representations of $SO(2n)$ into $U(n)$. This is obtained by grouping the gamma matrices as follows:
$$
\Gamma_{a}=\frac{1}{2}\left(\Gamma^{a}+i\Gamma^{a+n}\right)
$$
$$
\Gamma_{\bar{a}}=\frac{1}{2}\left(\Gamma^{a}-i\Gamma^{a+n}\right)
$$
Note that this new index $a$ labels the fundamental representations of the $U(n)$ subgroup of $SO(2n)$, while $\bar{a}$ labels the anti-fundamental representation. Raising the anti-fundamental $U(n)$ index by the $U(n)$ metric we have the following algebra:
$$
\{\Gamma_{a},\Gamma_{b}\}=0,\,\,\,\,\,\,\{\Gamma^{a},\Gamma^{b}\}=0,\,\,\,\,\,\,\{\Gamma_{a},\Gamma^{b}\}=\delta_{a}^{b}
$$
This is the well know algebra of a fermionic quantum oscillator. A representation is built by fixing a ground state that is annihilated by all the $\Gamma^{a}=\Gamma_{\bar{a}}$ annihilation operators, and others states can then be obtained by exciting this ground state with the raising operators $\Gamma_{a}$.
Example, for $d=10$, we have $n=5$, then we have the ground state 
$$
|0\rangle=|-----\rangle
$$
that is annihilated by all the $\Gamma^{a}$, and for instance: 
$$
\Gamma^{3}|-----\rangle=|--+--\rangle
$$
The charge conjugation matrix $C$ will be the one that conjugate all the $U(1)$ charges, and then switching fundamental representations to the anti-fundamental and vice versa:
$$
C|-----\rangle=|+++++\rangle
$$
while $C\Gamma_{a}=\pm\Gamma^{a}C$ and $C\Gamma^{a}=\pm\Gamma_{a}C$. Explicitly, this matrix can be written as 
$$
C=(\Gamma)\Gamma^{a+1}...\Gamma^{a+n}
$$
Where now, this indices are the $m$ kind of index, the $SO(2n)$ index. The $\Gamma$ inside the parenthesis is optional, and it is fixed if we fixed the signs of $C\Gamma_{a}=\pm\Gamma^{a}C$ and $C\Gamma^{a}=\pm\Gamma_{a}C$
In this notation, chirality is simply defined by the number of $+$ signs. If it is even, it is called Weyl or Chiral, if it is odd, it is called anti-Weyl or anti-Chiral. Then you see that depending on the dimension, the charge conjugation matrix can switch the chirality or not. More precisely, if $n$ is even then the chirality is preserved by the charge conjugation matrix, while if it is odd it will switch the chirality.
There are maybe some signs that I am missing here, but the idea is this.
