The organizational structure in MTW actually illustrates why there are 20 degrees of freedom and 21 unique (up to sign) nonzero components in the Riemann.
MTW start with the Riemann ##R^{\hat{a}\hat{b}}_{\hat{c}\hat{d}}## as per MTW's exercise $14.14, Note that this is presumed to be in an orthonormal basis, MTW uses the "hats" to convey this important information symbolically but I'll mention that as people not familiar with the books conventions wouldn't necessarily know that.
We start out with considering the "left half", ##R^{\hat{a} \hat{b}} ## and the "right half" ##R_{\hat{c}\hat{d}}##. Which are equivalent as far as numerical component values go, because the basis inner product ##n_{\hat{a}\hat{b}}## is diag(-1,1,1,1), and we multiply by it twice to raise and lower both indices which gives us an identity matrix for the numerical values of the components.
It's unclear to me why MTW uses ##R^{\hat{a}\hat{b}}{}_{\hat{c}\hat{d}}## rather than ##R_{\hat{a}\hat{b}\hat{c}\hat{d}}## though, to be honest.
Each half has two components belonging to the set (t,x,y,z) or (0,1,2,3) depending on your choice of notation (MTW uses the numeric indices). So a pair has 16 possible values.
So far, we've simply reorganized the 4x4x4x4 Rimenn as a 16x16 matrix, we haven't gained much yet.
But terms in each half with a repeated index are zero due to symmetry, so that leaves only 12 nonzero components. And (Anti)symmetry lowers this to six.
So, we've gone from a 4x4x4x4 display with 256 components to a 6x6 display with only 36 components. Which is a lot of progress, but we're not done yet.
In this 6x6 display, there are 6 diagonal elements and 30 off-diagonal elements. The off-diagonal elements are symmetrical (up to sign?). So that's 6+15=21 unique components.
The 21 components are not all independent, though - there's an identiy that lowers this from 21 to 20.
The 6x6 matrix form is further split up into 4 3x3 matrices. (That's back to the 36).
It looks like
$$\begin{bmatrix}E & H \\-H^T & F\\ \end{bmatrix}$$
The electrogravitic part is the "upper left" of the 4 3x3 matrices in MTW's scheme, each 3x3 matrix in the electrogravitic part has component pairs like tx, ty, and tz, pairs that include a time coordinate.
For a flow/vector field that points strictly along the coordinate time, the electrogravitic part is the only source of geodesic deviation, i.e. tidal forces. So a static frame for a static arrangment of masses only has the E part.
The split between E, H, and F is based on the number of "time" components. E has two, H has one, F has zero.
In Bel's original, F is named the topogravitic part, but the "modern" approach organizes things a bit differently as my recent reading mentioned. But I am still used to the "old" split.
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