Consider a tensor, $T$ of rank $(r,s)$ over a supermanifold, $M$ and take the supertrace over its indices $p$ and $q$ (DeWitt, p. 77, eq. 2.4.33):


As an example, the supertrace of a matrix (only defined when the indices are arranged from southwest to northeast or northwest to southeast) is given by: $$\text{str}({_iK^j})=(-1)^i{_iK^i}.$$

The other arrangement can be obtained by taking the supertransposition: $$_iL^{\sim j}=(-1)^{i(j+i)}{^jL_i}.$$

These definitions give rise to a supertrace that is oblivious to supertransposition.

Needless to say, a connection is not a tensor. However, one can still contract the indices of a connection by using the graded Jacobi formula (DeWitt, p. 34, 43, 113): $$(-1)^i\Gamma^i_{\,\,ij}=\text{ln}(g)_{,j},$$

where $g\equiv|\text{sdet}({_ig_j})|.$

How does one construct the supertraceless portion of a connection? This is shown on page 4 of George: Projective Connections and Schwarzian Derivatives for Supermanifolds, and Batalin-Vilkovisky Operators (arXiv:0909.5419v1). Unfortunately, I cannot follow his work or understand his conventions.

A first guess would be the following, but one runs into certain terms that appear in the coordinate transformation that do not belong there:

$$\Pi^a_{\,\,bc}\equiv \Gamma^a_{\,\,bc}-\frac{1}{m-n+1}\big((-1)^e\Gamma^e_{\,\,ec}\,{_b\delta^a}+(-1)^e\Gamma^e_{\,\,eb}\,{_c\delta^a}\big).$$

If the torsion tensor vanishes and we are making use of the coordinate basis, then the connection obeys the following rule: $$\Gamma^i_{\,\,jk}=(-1)^{jk}\Gamma^i_{\,\,kj}.$$

Taking the supertrace over $i$ and $j$ of the above, we have: $$(-1)^f\Gamma^f_{\,\,fk}=(-1)^{fk}(-1)^{f(1+k)}\Gamma^f_{\,\,kf}=(-1)^f\Gamma^f_{\,\,kf}=(-1)^{f(1+k)}\Gamma^f_{\,\,fk}.$$

Where am I messing up? I am using the conventions found in DeWitt Supermanifolds (2nd edition).


The usual tracing rules of DeWitt do not apply to the connection, and in general, they do not apply to any object that is not tensorial. The tracing of indices of nontensorial objects is carried out by contracting and adjoining a parity factor that reflects the parity of the contracted index: $$\Gamma^a_{\,\,bc}\rightarrow (-1)^d\Gamma^d_{\,\,bd}.$$

Note that we did not take into account that there were indices between $a$ and $c$. This can also be appreciated by deriving the graded Jacobi formula.

Below is the supertraceless sector of a connection:

$${\Pi}{^a_{\,\,bc}}\equiv \Gamma{^a_{\,\,bc}}-\frac{1}{m-n+1}\big({^a\delta_b}(-1)^e\Gamma{^e_{\,\,ec}}+{^a\delta_c}(-1)^{e+bc}\Gamma{^e_{\,\,eb}}\big).$$

Let $D=(m-n+1)^{-1}$. First, the supertrace over $a$ and $b$: \begin{align} &(-1)^f\Gamma{^f_{\,\,fc}}-D\big((-1)^f{^f\delta_f}(-1)^e\Gamma{^e_{\,\,ec}}+(-1)^{f}{^f\delta_c}(-1)^{e+fc}\Gamma{^e_{\,\,ef}}\big)\nonumber\\ &=(-1)^f\Gamma{^f_{\,\,fc}}-D\big((-1)^e\Gamma{^e_{\,\,ec}}(m-n)+(-1)^{e}\Gamma{^e_{\,\,ec}}\big)=0.\nonumber \end{align}

Supertrace over $a$ and $c$: \begin{align} &(-1)^{f}\Gamma{^f_{\,\,bf}}-D\big((-1)^{f}{^f\delta_b}(-1)^e\Gamma{^e_{\,\,ef}}+(-1)^f{^f\delta_f}(-1)^{e+bf}\Gamma{^e_{\,\,eb}}\big)\nonumber\\ &=(-1)^{bf+f}\Gamma{^f_{\,\,fb}}-D\big((-1)^{bf}(-1)^{e}\Gamma{^e_{\,\,ef}}{^f\delta_b}+(-1)^{bf}(-1)^e\Gamma{^e_{\,\,eb}}(-1)^f{^f\delta_f}\big)\nonumber\\ &=(-1)^{bf}\big((-1)^f\Gamma{^f_{\,\,fb}}-D\big((-1)^e\Gamma{^e_{\,\,ef}}{^f\delta_b}+(-1)^e\Gamma{^e_{\,\,eb}}(m-n)\big)\big)=0.\nonumber \end{align}

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