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$A$ is a tensor field, $\nabla A(X,...)=(\nabla_XA)(...)$. Besides, $R$ is curvature operator and $$ Rm(X,Y,Z,W)= \langle R(X,Y)Z,W\rangle $$ and $$ \Delta A = g^{ij}(\nabla _{X_i}\nabla _{X_j}A - \nabla_{\nabla_{X_i}X_j}A) $$ For any tensor field $A,B$, $A*B$ is linear combination of tensor fields, each formed by starting with the tensor field $A\otimes B$, using metric to switch the type of components or to contractions. Then, how to show $$ \nabla (\Delta A) -\Delta(\nabla A)=\nabla Rm *A+Rm*\nabla A ~~~? \tag{1} $$

PS: I know how to show $R(\cdot,\cdot)A=A*Rm$, for example $A=A_{ij}dx^i\otimes dx^j$, I can get $$ [R(\cdot,\cdot)A]_{abcd}=-g^{lk}R_{abck}A_{ld}-g^{lk}R_{abdk}A_{cl} $$ therefore, there is $$ R(\cdot,\cdot)A=A*Rm $$ But I fail to show (1).

Enhao Lan
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    After I posted my answer, I noticed that your question was essentially a duplicate of this one. Anthony's answer is equivalent to what I've tried to express, but, perhaps, is more accessible for you. – Yuri Vyatkin Sep 25 '21 at 13:30

1 Answers1

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Firstly, let me restate the definition of the notation $A*B$ as a linear combination of various compatible contractions of $A \otimes B$. This notation is very permissive and absorbing, e.g.: $$ - A*B \equiv A*B \tag{1} $$ and $$ A*B + A*B \equiv A*B \tag{2} $$ where by $\equiv$ here I mean "can be renamed as".

Secondly, for the sake of easier calculations, let me switch to the abstract index notation (see e.g. R.Wald, General relativity, for an accessible introduction).

Thus, the Ricci calculus is presented by the following identities. $$ 2 \nabla_{[a} \nabla_{b]} X^c = R_{a b}{}^{c}{}_{d} X^d \tag{3} $$ and $$ 2 \nabla_{[a} \nabla_{b]} \omega_d = - R_{a b}{}^{c}{}_{d} \omega_c \tag{4} $$

You may be wondering about the sign conventions for the curvature, but for our needs it does not really matter, as we are going to rewrite the above equations as $$ \nabla_a \nabla_b X^c = \nabla_b \nabla_a X^c + (R*X)_{a b}{}^c \tag{5} $$ and $$ \nabla_a \nabla_b \omega_c = \nabla_b \nabla_a \omega_c + (R*\omega)_{a b c} \tag{6} $$

Since any tensor $A$ is can be presented as a linear combination of simple terms, that is tensor products of some amount of vectors and covectors, by the virtue of the linearity of the covariant derivative and using the Leibniz rule, we can see that $$ \nabla_a \nabla_b A = \nabla_b \nabla_a A + (R*A)_{a b} \tag{7} $$ for any tensor $A$ (this is the Ricci identities in disguise).

Now, down to the calculation. $$ (\nabla \Delta A)_a - (\Delta \nabla A)_a = \nabla_a \Delta A - \Delta \nabla_a A = \\ \nabla_a (g^{b c} \nabla_b \nabla_c A) - g^{b c} \nabla_b \nabla_c \nabla_a A = \\ g^{b c} ( \nabla_a \nabla_b \nabla_c A - \nabla_b \nabla_c \nabla_a A ) \equiv \\ g^{b c} ( \nabla_b \nabla_a \nabla_c A + (R*\nabla A)_{b a c} - \nabla_b \nabla_c \nabla_a A ) \equiv \\ g^{b c} ( \nabla_b \nabla_a \nabla_c A + (R*\nabla A)_{b a c} - \nabla_b ( \nabla_a \nabla_c A + (R*A)_{a c}) ) \equiv \\ g^{b c} ( \nabla_b \nabla_a \nabla_c A + (R*\nabla A)_{b a c} - \nabla_b \nabla_a \nabla_c A + (\nabla R*A)_{b a c} + (R * \nabla A)_{b a c} ) \equiv \\ (\nabla R*A)_a + (R*\nabla A)_a $$

Let me emphasize, that the $*$-notation should be understood as $$ \nabla \Delta A - \Delta \nabla A \equiv \nabla R*A + R*\nabla A $$

Yuri Vyatkin
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  • Nice, but I think you forgot an $RA$ term in the 2nd line, because the Laplacian also has a first order term, and the difference between the first order term coming from the derivative of the Laplacian of $A$, and that coming from the Laplacian of the derivative of $A$, should give you an $RA$ term (unless I am mistaken somewhere). – Malkoun Sep 25 '21 at 13:28
  • @Malkoun Thank you. I think that I just expand the definition there. This is the so-called "connection Laplacian", which in the abstract index notation is defined as $\Delta A := g^{a b} (\nabla \nabla A)_{a b}$, where the indices, coming from tensor $A$ are suppressed (not shown). For a discussion of the second covariant derivative $\nabla \nabla A = \nabla^2 A$ please check this Wikipedia article. – Yuri Vyatkin Sep 25 '21 at 13:49
  • @Malkoun Also, I have to mention that it is an unfortunate tradition to write $\nabla_a \nabla_b A$ instead of $(\nabla \nabla A)_{a b}$, but I assume that it is well known. Actually, this is why I use the parentheses to clarify the meaning of the indices, like in the first line of the calculation. – Yuri Vyatkin Sep 25 '21 at 13:54
  • I see. I must admit it is a somewhat confusing abuse of notation. Is this part of the abstract index notation? I should know these things better (but I haven't computed with curvature in a while). – Malkoun Sep 25 '21 at 14:01
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    @Malkoun yes, this is one of the quirks that the abstract index notation has inherited from the classical Ricci calculus (in order to keep the equations look the same). Using the parentheses is a pedantic way to mitigate this confusion. Reading Wald's book really helps. You could also compare my answer with the Antony's one. – Yuri Vyatkin Sep 25 '21 at 14:06
  • Thank you so much. It makes sense to me now. I should have spent more time in the past understanding the conventions for second order covariant derivatives :). But I guess it is better late than never. Thanks! – Malkoun Sep 25 '21 at 14:29
  • Shouldn't $\nabla_a \nabla_b A=(\nabla \nabla A){ba} $ to mimic $\nabla_a\nabla_b A=A{|b|a}$? – ContraKinta Sep 25 '21 at 14:48
  • @ContraKinta sorry, I don't understand what you mean. You can check out how I use the notation from this answer, for instance. – Yuri Vyatkin Sep 25 '21 at 15:21
  • I'm just curious about the order of the slots. I interpret your notation $(\nabla \nabla A){ab}$ as $A$ is a tensor with an unknown number of suppressed indices. Each time you form a covariant derivative you add another lowered index. First you form the covariant derivative (adding the slot $A{|b}$) and then you form the covariant derivative adding slot $A_{|b|a}$. For a torsion free connection we can express the lack of symmetry of the tensor $A^j_{|b|a}$ in the indices $b$ and $a$ as $(\nabla_a\nabla_b-\nabla_b\nabla_a)A=A^j_{|b|a}-A^j_{|a|b}=K_{l,,ba}^{,,j}A^l$ – ContraKinta Sep 25 '21 at 16:58
  • @ContraKinta Ah, I see. Thank you for the details. Simply put, you are using a different set of conventions for the index placement. There is a variety of traditions in differential geometry and mathematical physics in this regard. I stick to the convention that $\nabla_a \nabla_b A_{\dots}$ is a "traditional" notation for the second covariant derivative $(\nabla \nabla A)_{a b \dots}$, where $\dots$ denote the suppressed indices of $A$. Beware that there is much more sloppiness in this notation. In my thesis I made a humble attempt to clarify this a little further. – Yuri Vyatkin Sep 26 '21 at 01:52
  • Thanks your help. Does $\nabla_a \nabla_b X^c$ mean $\nabla_a \nabla_b X^c dx^a \otimes dx^b \otimes \partial_c$ ? – Enhao Lan Sep 28 '21 at 05:35
  • @lanse7pty no, the abstract index notation works differently. It means the tensor (field) $\nabla \nabla X$, where the indices show which tensor space (or bundle) it is taken from. In contrast, the notation $\nabla_a \nabla_b X^c dx^a \otimes dx^b \otimes \partial_c$ assumes a choice of coordinates, and $\nabla_a \nabla_b X^c$ denotes the components of this tensor in the chosen coordinates. Please take a look at this answer. – Yuri Vyatkin Sep 28 '21 at 05:47
  • @lanse7pty This is why I write $\nabla_a \nabla_b X^c = (\nabla \nabla X)_{a b}{}^{c}$ interchangeably, and even can drop (suppress) the indices completely, when the space is clear from the context. – Yuri Vyatkin Sep 28 '21 at 05:54
  • I have read your other answer before I ask question. But I misunderstood you. Besides, I use to the definition $R(X,Y)Z = \nabla _Y\nabla _X Z - \nabla_X \nabla _Y Z$. And only from this point , I can get (5). I don't know how to from the abstract index notation to get (5). Could you talk it ? Thanks. – Enhao Lan Sep 28 '21 at 12:05
  • @lanse7pty I will try to give you more explanations in my answer to your another related question which I am planning to write. – Yuri Vyatkin Sep 28 '21 at 12:12
  • I like you. Thanks very much. – Enhao Lan Sep 28 '21 at 12:13
  • I have get (5) from the traditional notation, namely, $R(X,Y)Z = \nabla _Y\nabla _X Z - \nabla_X \nabla _Y Z$. And I think I understand abstract index notation by reading it in Wiki. But I can't get its advantage. In contrast, the trace and Einstein summation are confused in my view. Besides, I think an important thing for using abstract index notation is that the order of abstract indices is fixed. Thanks again. – Enhao Lan Sep 29 '21 at 14:47
  • @lanse7pty The Riemann curvature operator is a tensor, defined as $R(X,Y)Z = \nabla Y\nabla _X Z - \nabla_X \nabla _Y Z - \nabla{[ X, Y]} Z$. Your formula is valid only for coordinate vector fields. Because it is a tensor (field), we can write it as $R_{a b}{}^{c}{}_{d}$ where the indices only indicate the space, this field is a section of. This is why they are called "abstract" in this case. The trace and the Einstein summation appear (in disguise) as contraction, e.g. $\omega_a X^a$ means $\omega(X)$. – Yuri Vyatkin Sep 29 '21 at 23:19
  • @lanse7pty There are advantages and disadvantages in using the abstract index notation, as any other notation, in general. We make choices and trade offs. For me, using abstract indices, saves a lot of space and gives a lot of freedom for being less careful, but still being correct. You just need to grasp their value. Also, Anthony has already given an answer in a more traditional way. – Yuri Vyatkin Sep 29 '21 at 23:26