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On pg. 71 of Dalarsson’s Tensors, Relativity and Cosmology

10.4 Curl of Vectors

In an arbitrary $N$-dimenstional metric space the curl of a vector function $A_m$ is a second-order covariant tensor $F_{mn}$, defined by the expression $$ F_{mn} = D_m A_n - D_n A_m ~~~(m,n = 1,2,\dots,N). \tag{10.19} $$ In the special case of a three-dimensional metric space, it is possible to construtct a contravariant vector $C^k$ related to the tensor (10.19) using the three-dimensional totally antisymmetric Ricci tensor $\varepsilon^{kmn}$, defined according to the general definition (6.23) as follows: $$ \varepsilon^{kmn} = \frac1{\sqrt g} e^{kmn}~~~(k,m,n = 1,2,3).\tag{10.20} $$ Thus the curl in the three-dimensional metric space is defined by $$ C^k = \frac12 \varepsilon^{kmn} F_{mn} = \varepsilon^{kmn} D_m A_n = \frac1{\sqrt g} e^{kmn} D_m A_n.\tag{10.21} $$

The author introduced the definition of the curl of the vector as a tensor in (10.19) then, in (10.21) the $\tfrac 1 2F_{mn}$ is converted directly into $D_m A_n $ which is confusing for me. Doesn’t this substitution assumes that $D_m A_n $ is antisymmetric with respect to both of its indices? (Which has never been mentioned in the text). If anyone could help me out I’ll really appreciate it!

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  • $\begingroup$ 1. Have you tried writing out (10.19) in the form of partial derivatives and Christoffel symbols and checking whether it yields zero? 2. For the step in (10.21) that is confusing you, just substitute (10.19) and remember that $\epsilon$ is anti-symmetric. 3. Please do not post formulae as screenshots, but use MathJax instead. $\endgroup$ – ACuriousMind Apr 19 '19 at 23:03
  • $\begingroup$ Why would (10.19) be zero? It's not zero even when curvature is zero, that is, in flat space. There you have, for instance, $\partial_t A_x - \partial_x A_t$. What makes you think this is zero? $\endgroup$ – Avantgarde Apr 19 '19 at 23:08
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Any 2-tensor can be written as a sum of a symmetric and antisymmetric part, $$ \begin{align} A_{mn} &= A_{(mn)} + A_{[mn]} \text{ where}\\ A_{(mn)} &= \frac12 A_{mn} + \frac12 A_{nm},\\ A_{[mn]} &= \frac12 A_{mn} - \frac12 A_{nm}. \end{align} $$ One has general theorems that $$A^{(mn)} B_{[mn]} = A^{[mn]} B_{(mn)} = 0$$as can be discovered by simply writing them out: $$4A^{(mn)}B_{[mn]} = A^{mn} B_{mn} + A^{nm}B_{mn} - A^{mn}B_{nm} - A^{nm}B_{nm}$$ This may not look like it cancels at first but any index that appears in a contraction may be safely relabeled, and by carefully relabeling the latter two terms with $m \leftrightarrow n$ we would find that this is exactly $$4A^{(mn)}B_{[mn]} = A^{mn} B_{mn} + A^{nm}B_{mn} - A^{nm}B_{mn} - A^{mn}B_{mn},$$which obviously vanishes as it has the form $X + Y - Y - X = 0.$

As a consequence an inner product can always be reduced as $$A^{mn} B_{mn} = A^{(mn)}B_{(mn)} + A^{[mn]} B_{[mn]}.$$In the case of the orientation tensor, however, it is totally antisymmetric, having thus $$\varepsilon_{abc} = \varepsilon_{[ab]c} = \varepsilon_{a[bc]} = \varepsilon_{[a|b|c]},$$ using the pipe character $|$ to "interrupt" antisymmetry brackets. As a result when we try the same trick with $\varepsilon^{abc} D_b A_c$ we find directly $$ \begin{align} \varepsilon^{abc} D_b A_c &= \varepsilon^{a(bc)} D_{(b}A_{c)} + \varepsilon^{a[bc]} D_{[b} A_{c]}\\ &= 0 + \varepsilon^{abc} D_{[b} A_{c]}\\ &= \varepsilon^{abc} \left(\frac12 D_b A_c - \frac12 D_c A_b\right), \end{align} $$ which was to be proven.

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