Representation theory of (anti)self-dual tensors  I am using usual physics notations and I guess the physics motivations of this question are obvious. 
Let a basis of the $SO(n,m)$ Lie algebra be denoted by $S^{\mu \nu}$ and the Lie algebra be, $[S^{\mu \nu},S^{\lambda \rho}] = i(g^{\mu \rho}S^{\nu \lambda} + g^{\nu \lambda}S^{\mu \rho} - g^{\mu \lambda}S^{\nu \rho} - g^{\nu \rho}S^{\mu \lambda})$ where $g$ is the matrix $diag(-1,-1,..m-times..,-1,1,1,..n-times..,1)$. 


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*At least in the $m=1$ (Lorentz) case it is true that if one can find a set of matrices $\Gamma^\mu$ such that they satisfy the "corresponding" Clifford algebra, $[\Gamma^\mu, \Gamma ^\nu] = 2g^{\mu \nu}$ then a representation of the $SO(n,1)$ Lie algebra is given by $S^{\mu \nu} = \frac{i}{4}[\Gamma ^\mu , \Gamma ^\nu]$ 
Does the above construction work for arbitrary $m$ , especially $m = 0,2$ and what is the global understanding for why this should work? 

*The representation $S^{\mu \nu} = \frac{i}{4}[\Gamma ^\mu , \Gamma ^\nu]$ is how the 
$SO(n,1)$ Lie algebra acts on its spinorial representations. (..those representations whose weights are given by a $[\frac{n+1}{2}]$ tuple of $\pm \frac{1}{2}$..)   
What is the $m \neq 1$ generalization of the above? (...one case that I have often seen used are these two earlier questions of mine..)

*Now my main question is to understand how the above construction - at least for the most familiar case of $n=3, m=1$ - gives an alternative to using the language of tensors. 
Like to give probably the most used example - if $F_{\mu \nu}$ is a $4$-dimensional antisymmetric rank $2$ tensor then one defines the quantities $F_{\alpha \beta}$ and $\bar{F} _ {\dot{\alpha} \dot{\beta}}$ s.t $F_{\alpha \beta} = (S^{\mu \nu}) _ {\alpha \beta} F_{\mu \nu}$ and $\bar{F} _ {\dot{\alpha} \dot{\beta}} = (S^{\mu \nu})_{\dot{\alpha}  \dot{\beta}} F_{\mu \nu}$ 


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*What exactly is the group/representation theoretic meaning behind defining these $F_{\alpha \beta}$ and $\bar{F}_{\dot{\alpha} \dot{\beta}}$? 

*In what sense is knowing the $F_{\alpha \beta}$ and $\bar{F}_{\dot{\alpha} \dot{\beta}}$ equivalent to knowing the $F_{\mu \nu}$?

*In what sense is $F_{\alpha \beta}$ and $\bar{F}_{\dot{\alpha} \dot{\beta}}$ the self-dual ($\frac{1}{2}(F + *F)$) and the anti-self-dual ($\frac{1}{2}(F - *F)$) parts of the tensor $F$? 


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*I guess from here one can also explain why the self-dual part is thought to be in the $(1,0)$ representation and the anti-self-dual part is thought to be in the $(0,1)$ representation of either the $SL(2,\mathbb{C}) \times SL(2,\mathbb{C})$ (..isometry group of the complexified Minkowski space..) or of $SU(2) \times SU(2)$ (..in in Euclidean four dimensional space..) 

*Is there an analogue of the $F_{\alpha \beta}$ and $\bar{F}_{\dot{\alpha} \dot{\beta}}$ for general $SO(n,m)$ and arbitary dimension higher rank tensors ? 
 A: I think what is going on is the following.  I find it a bit difficult to wade through your notation, so I am going to use my own (or rather, the notation of Chapter 5 of Elements of Noncommutative Geometry, where I copied most of this from).
Edit: this got kind of long, so let me say what the point is.  You start with a real vector space $V$ with a symmetric bilinear form $g$ and construct the corresponding Clifford algebra $\mathrm{Cl}(V,g)$.  The construction that I detail below constructs an isomorphism of $\mathfrak{so}(V,g)$ with a certain subspace of "degree 2 elements" in $\mathrm{Cl}(V,g)$ (quotes because the Clifford algebra is not a graded algebra).  I believe that this is the source of the connection that you are trying to find.
Let $V$ be a finite-dimensional vector space over $\mathbb{R}$ equipped with a symmetric bilinear form $g$ (of any signature).  Edit: $g$ should be nondegenerate. 
It is standard (see for instance this question)
 that the Clifford algebra $\mathrm{Cl}(V,g)$ is isomorphic as a vector space to the exterior algebra $\Lambda(V)$.  In particular, looking just at the degree 2 part, there is an injective linear map $Q : \Lambda^2(V) \to \mathrm{Cl}(V,g)$ given by
$$ Q(v \wedge w) = \frac{1}{2} (vw - wv). $$
For brevity, let's denote $b = Q(v \wedge w) = \frac12(vw - wv) \in \mathrm{Cl}(V,g)$, and consider the operator $\mathrm{ad}_b = [b,-]$ on $\mathrm{Cl}(V,g)$.  In particular, consider what this operator does to a vector $x \in V \subseteq \mathrm{Cl}(V,g)$.
First, note that in the Clifford algebra we have $wv = - vw + 2g(v,w)$, so
$$ \mathrm{ad}_b(x) = \left[\frac12(vw - wv), x\right] = [vw, x]$$
since the scalar term commutes with $x$.
Then using the relation $[ab,c] = a(bc + cb) - (ac + ca)b$, (which holds trivially in any associative algebra), we get
\begin{align*}
  \mathrm{ad}_b(x) & = [vw,x] \\\
  & = v(wx + xw) - (vx + xv)w \\\
  & = 2g(w,x)v - 2g(v,x)w.
\end{align*}
In other words, the operator $\mathrm{ad}_b$ preserves the subspace $V \subseteq \mathrm{Cl}(V,g)$.
Next, we check that this operator is actually in $\mathfrak{so}(V,g)$.  From our calculations above, we have for any $x,y \in V$:
$$ g(y,[b,x]) = 2g(w,x)g(y,v) - 2g(v,x)(y,w) = -g([b,y],x). $$
Now the map $b \mapsto \mathrm{ad}_b$ is injective, so for dimension reasons this gives that
$$ \{ \mathrm{ad}_b|_V : b \in Q(\Lambda^2(V)) \} = \mathfrak{so}(V,g). $$
This was all independent of signature, so it works for any symmetric bilinear form.
Sorry I couldn't dovetail this all a little better with your notation!
Edit: injectivity of the map $b \mapsto \mathrm{ad}_b$ is not totally trivial, and may only hold for certain signatures.  It requires you to analyze the intersection of the even subalgebra with the center of $\mathrm{Cl}(V,g)$.
Here is how the argument should go: If $\mathrm{ad}_b$ is zero, then $b$ is even and central.  Then you need to be able to conclude that $b$ is scalar, and then that $b = 0$.  It requires a case-by-case analysis (probably there is a better way, but I don't see it right now) to determine when the even part of the center consists just of scalars.  Then you should be able to use the trace on the Clifford algebra to conclude that $b=0$.
Further edit: I think my the argument that $b \mapsto \mathrm{ad}_b$ is injective actually does work in all signatures.
As I noted above, if $\mathrm{ad}_b = 0$ then $b$ is an even, central element of the Clifford algebra.  We want to conclude that $b$ must be scalar.  In most signatures, the Clifford algebra is simple, so the center consists just of scalars, so we're done.  But in the two case where the Clifford algebra is not simple, the even subalgebra is simple, so a central even element must be scalar.
Then we want to show that if $\mathrm{ad}_b = 0$ then we must have $b=0$ (only for $b$ which is a linear combination of terms of the form $\frac12 (vw - wv)$).  There is a unique trace $\tau$ on $\mathrm{Cl}(V,g)$ which vanishes on the odd part and satisfies $\tau(1)=1$; this is given by taking the coefficient of the identity element in a basis for the Clifford algebra consisting of increasing products of elements of a basis for $V$ (and this is independent of the basis).  But the trace vanishes on our element $b$, since it is given to us as a sum of commutators.  Thus $b=0$.
