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In tensor calculus I am reading two different books. Both have different notations. $$D_u$$ as well as $$\nabla_u$$. In the context of second covariant differentiation for some tensor of rank (2,0) $$T^{nm}$$ the second covariant derivative is denoted $$\nabla_a\nabla_bT^{nm}$$ but what does $$\nabla_{\nabla_ab}$$ mean. Is it the same thing? How would I compute $$\nabla_{\nabla_ab}T^{nm}$$? Same as $$\nabla_a\nabla_bT^{nm}$$?

aygx
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  • Isn’t one differentiating $T$ just once? – Ted Shifrin Oct 06 '21 at 01:46
  • Yes $$\nabla_bT^{ij}$$ is just differentiating T once. – aygx Oct 06 '21 at 01:59
  • Im curious as to see if $$\nabla_{\nabla_ab}$$=$$\nabla_a\nabla_b$$ – aygx Oct 06 '21 at 02:00
  • So you didn’t think about my point. The first differentiates $T$ once, not twice. – Ted Shifrin Oct 06 '21 at 03:07
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    Maybe your second book is expressing the second covariant derivative in a coordinate-free manner using the linear map $\frak{X}$(M)$\times\frak{X}$(M)$\to$$\frak{X}$(M) denoted by $\nabla:(X,Y)\to\nabla_X Y$. In that case $\nabla^2_{yx}Z=\nabla_y\nabla_xZ-\nabla_{\nabla_yx}Z$. In ain this is equivalent to $y^mx^k\nabla_m\nabla_kZ^j$ – ContraKinta Oct 06 '21 at 03:37
  • Is $$Z^j$$ a tensor? – aygx Oct 06 '21 at 10:40
  • Yes $Z^j$ is a type $(1,0)$-tensor(field). The second covariant derivative of $Z^j$, $\nabla_m\nabla_kZ^j$, is a type $(1,2)$-tensor(field). The contraction $y^mx^k\nabla_m\nabla_kZ^j$ is a type $(1,0)$-tensor(field). – ContraKinta Oct 06 '21 at 14:09
  • so $$Z^i$$ is a rank 1 tensor. The expression $$\nabla^2_{xy}$$ is the second covariant derivative right? – aygx Oct 06 '21 at 14:39
  • Since this is true, how would I compute $$\nabla_a(\nabla_bT^iG^j)$$? – aygx Oct 06 '21 at 14:51
  • If $a$ and $b$ are indices you can use the product rule.

    But please note the difference between $\nabla^2_{XY}Z=(\nabla_X\nabla_Y-\nabla_{\nabla_X}Y)Z$ where $X,Y,Z\in\mathscr{X}(M)$ and $\nabla_a\nabla_bZ^j\in \mathscr{T}^1_2$ where $a,b$ and $j$ are abstract indices. Pick one meaning of "second covariant derivative".

     – ContraKinta Oct 06 '21 at 20:48
  • So would this expression be $$\nabla_a(\nabla_bT^iG^j)=\nabla_b T^i\nabla_aG^j+G^i\nabla_a\nabla_bT^i$$? – aygx Oct 06 '21 at 21:41
  • Yes. Except $G^j$ not $G^i$ in the last term. And note that this is not the same thing as $\nabla_a\nabla_b(T^iG^j)$ – ContraKinta Oct 06 '21 at 21:54
  • What would It then be for $$\nabla_a\nabla_b(T^iG_m)$$ and $$\nabla_a\nabla_bT^iG_m$$? Would I just act with $$\nabla_b$$? – aygx Oct 06 '21 at 23:05

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