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Support Theorems and an Injectivity Result for Integral Moments of a Symmetric m-Tensor Field

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Abstract

In this work, we show an injectivity result and support theorems for integral moments of a symmetric m-tensor field on a simple, real analytic, Riemannian manifold. Integral moments of symmetric m-tensor fields were first introduced by Sharafutdinov. First we generalize a Helgason type support theorem proven by Krishnan and Stefanov (Inverse Probl Imaging 3(3):453–464, 2009). We use this extended result along with the first integral moments of a symmetric m-tensor field to prove the aforementioned results.

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Acknowledgements

We would like to thank Prof. Vladimir Sharafutdinov for suggesting this problem. Besides, we would also like to express our sincere gratitude to Prof. Todd Quinto and Prof. Venky Krishnan for several hours of fruitful discussions. The authors benefited from the support of the Airbus Group Corporate Foundation Chair “Mathematics of Complex Systems” established at TIFR Centre for Applicable Mathematics and TIFR International Centre for Theoretical Sciences, Bangalore, India. Finally, we would like to thank the referee for their helpful comments and suggestions.

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Correspondence to Rohit Kumar Mishra.

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Communicated by Alex Iosevich.

Appendix

Appendix

Proof of Lemma 1

First, let us recall that for a symmetric \((m-1)\)-tensor field v,

$$\begin{aligned} (\mathrm {d}v)_{i_1\dots i_m} = \sigma (i_1,\ldots ,i_m)\left( \frac{\partial v_{i_1\ldots i_{m-1}}}{\partial x^{i_m}}-\sum _{l=1}^{m-1}\Gamma _{i_m i_l}^pv_{i_1,\ldots i_{l-1} p i_{l+1}\ldots i_{m-1}}\right) . \end{aligned}$$

The idea here is to use an inductive argument for \(0 \le k \le m\). We start by showing the result for \(k=0,1,\) and then for general \(k\le m\).

$$\begin{aligned} (\mathrm {d}v)_{n\ldots n }&=\frac{\partial v_{n\ldots n}}{\partial x^{n}},\quad \text{ for } k=0\\ (\mathrm {d}v)_{n\ldots n i}&= \frac{m-1}{m}\frac{\partial v_{n\ldots n i }}{\partial x^n}-\frac{2(m-1)}{m}\Gamma _{in}^pv_{n\ldots n p}+\frac{1}{m}\frac{\partial v_{n\ldots n }}{\partial x^i},\quad \text{ for } k=1. \end{aligned}$$

From this, we see that the result is true for \(k=0\) and 1. Now, we are going to prove that the result is also true for \(k\le m\). Consider

$$\begin{aligned} (\mathrm {d}v)_{n\dots n i_k\ldots i_1}&= \sigma (n, \ldots n ,i_k ,\ldots , i_1)\\&\left( \frac{\partial v_{n\ldots n i_k\dots i_2}}{\partial x^{i_{1}}}- \sum _{l=2}^{k} \Gamma _{i_li_{1}}^p v_{n\ldots n i_k\dots {\hat{i_l}} \ldots i_2p}- (m-k)\Gamma _{n i_{1}}^pv_{n\ldots n i_k\dots i_2p} \right) \\&= J-J^1_{k} - (m-k)J^2_{k} \end{aligned}$$

where

$$\begin{aligned} J&=\sigma (n, \ldots n ,i_k ,\ldots , i_1)\left( \frac{\partial v_{n\dots n i_k\ldots i_2}}{\partial x^{i_{1}}}\right) ,\\ J^1_{k}&= \sigma (n, \ldots n ,i_k ,\ldots , i_1)\left( \sum _{l=2}^{k} \Gamma _{i_li_{1}}^p v_{n\dots n i_k\dots {\hat{i_l}} \ldots i_2p}\right) ,\\ J^2_{k}&= \sigma (n, \ldots n ,i_k ,\ldots , i_1)\left( \Gamma _{n i_{1}}^pv_{n\ldots n i_k\dots i_2p}\right) . \end{aligned}$$

Now, we will simplyfy each of the above terms one by one. Consider

$$\begin{aligned} J&= \sigma (n, \ldots n ,i_k ,\ldots , i_1)\left( \frac{\partial v_{n\ldots n i_k\ldots i_2}}{\partial x^{i_1}}\right) \\&= \frac{\sigma (n, \ldots n ,i_k ,\dots , i_2)}{m}\left( \frac{\partial v_{n\ldots n i_k\dots i_2}}{\partial x^{i_1}} + (m-1)\frac{\partial v_{n\ldots n i_k\ldots i_1}}{\partial x^{n}}\right) \\&= \frac{1}{m}\frac{\partial v_{n\ldots n i_k\ldots i_2}}{\partial x^{i_1}} + \frac{\sigma (n, \ldots n ,i_k ,\ldots , i_{3})}{m}\left( \frac{\partial v_{n\ldots n i_k\ldots i_{3}i_{1}}}{\partial x^{i_{2}}}+(m-2)\frac{\partial v_{n\dots n i_k\dots i_{1}}}{\partial x^{n}}\right) \\&= \frac{1}{m}\sum _{l=1}^{k}\frac{\partial v_{n\ldots n i_k\ldots i_{l-1} i_{l+1}\dots i_{1}}}{\partial x^{i_l}}+ \frac{m-k}{m}\frac{\partial v_{n\dots n i_k\ldots i_{1}}}{\partial x^{n}},\quad \text { (repeating similar arguments)}.\\ J^2_{k}&= \sigma (n, \dots n ,i_k ,\ldots , i_1)\left( \Gamma _{n i_{1}}^pv_{n\dots n i_k\dots i_2p}\right) \\&= \frac{\sigma (n, \ldots n ,i_k ,\ldots , i_2)}{m}\left( 2\Gamma _{n i_{1}}^pv_{n\ldots n i_k\ldots i_2p}+ (m-2) \Gamma _{n i_{2}}^pv_{n\ldots n i_k\ldots i_3 i_1p}\right) \\&= \frac{2\sigma (n, \ldots n ,i_k ,\ldots , i_3)}{m(m-1)}\left( \Gamma _{i_2 i_{1}}^pv_{n\ldots n i_k\ldots i_3p}+(m-2)\Gamma _{ni_{1}}^pv_{n\dots n i_k\ldots i_2p}\right) \\&\quad + \frac{(m-2)\sigma (n, \ldots n ,i_k ,\ldots , i_3)}{m(m-1)}\left( 2 \Gamma _{n i_{2}}^pv_{n\ldots n i_k\ldots i_3 i_1p} + (m-3)\Gamma _{n i_{3}}^pv_{n\ldots n i_k\ldots i_4i_2 i_1p}\right) \\&= \frac{2}{m(m-1)}\Gamma _{i_2 i_{1}}^pv_{n\ldots n i_k\ldots i_3p}+\frac{2(m-2)\sigma (n, \ldots n ,i_k ,\ldots , i_3)}{m(m-1)}\sum _{q=1}^2\Gamma _{n i_{q}}^pv_{n\ldots n i_k\dots i_3 {\hat{i_q}}i_1p}\\&\quad + \frac{(m-3)(m-2)\sigma (n, \ldots n ,i_k ,\dots , i_3)}{m(m-1)}\Gamma _{n i_{3}}^pv_{n\ldots n i_k\ldots i_2 i_1p}\\&= \frac{2}{m(m-1)}\sum _{q,r=1,q\ne r}^3\Gamma _{i_q i_{r}}^pv_{n\ldots n i_k\ldots i_4{\hat{i_q}}{\hat{i_r}}i_1p}\\&\quad +\frac{2(m-3)\sigma (n, \ldots n ,i_k ,\dots , i_4)}{m(m-1)}\sum _{q=1}^3\Gamma _{n i_{q}}^pv_{n\ldots n i_k\dots i_4 {\hat{i_q}} i_1p}\\&\quad + \frac{(m-4)(m-3)\sigma (n, \dots n ,i_k ,\ldots , i_4)}{m(m-1)}\Gamma _{n i_{4}}^pv_{n\dots n i_k\dots i_5i_3 i_2 i_1p},\\&= \frac{2(k-1)}{m(m-1)}\sum _{q,r =1, q\ne r}^k\Gamma _{i_qi_{r}}^p v_{n\dots n i_k\ldots {\hat{i_q}}\ldots {\hat{i_r}}\ldots i_1p}+\frac{2(m-k)}{m(m-1)}\sum _{q=1}^k\Gamma _{ni_q}^p v_{n\ldots n i_k\ldots {\hat{i_q}} \ldots i_1p} \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \text{(repeating } \text{ similar } \text{ calculation } (k-3) \text{-times) }.\\ J^1_{k}&= \sigma (n, \ldots n ,i_k ,\ldots , i_1)\left( \sum _{l=2}^{k} \Gamma _{i_li_{1}}^p v_{n\ldots n i_k\ldots {\hat{i_l}} \ldots i_2p}\right) \\&= \frac{\sigma (n, \dots n ,i_k ,\ldots , i_2)}{m}\left( 2\sum _{l=2}^{k} \Gamma _{i_li_{1}}^p v_{n\ldots n i_k\dots {\hat{i_l}} \dots i_2p}+(m-2)\sum _{l=3}^{k} \Gamma _{i_li_2}^p v_{n\ldots n i_k\dots {\hat{i_l}} \ldots i_3i_1p}\right. \\&\quad +\left. (m-2) \Gamma _{i_3i_{2}}^p v_{n\ldots n i_k\ldots i_4i_1p}\right) \\&= \frac{\sigma (n, \ldots n ,i_k ,\ldots , i_3)}{m(m-1)}\left\{ 2(k-1) \Gamma _{i_2i_{1}}^p v_{n\ldots n i_k\ldots i_3p}+(m-2)\left( 2\sum _{l=3}^{k} \Gamma _{i_li_{1}}^p v_{n\ldots n i_k\dots {\hat{i_l}} \ldots i_2p}\right. \right. \\&\quad +2\Gamma _{i_3i_{1}}^p v_{n\ldots n i_k \ldots i_4 i_2p}+ 2\sum _{l=3}^{k} \Gamma _{i_li_2}^p v_{n\ldots n i_k\ldots {\hat{i_l}} \dots i_3i_1p}+(m-3)\sum _{l=4}^{k} \Gamma _{i_li_3}^p v_{n\ldots n i_k\dots {\hat{i_l}} \ldots i_4i_2 i_1p}\\&\quad \left. + (m-3) \Gamma _{i_3i_4}^p v_{n\ldots n i_k\ldots i_5i_2i_1p}+2 \Gamma _{i_3i_{2}}^p v_{n\ldots n i_k\ldots i_4i_1p}+(m-3) \Gamma _{i_3i_{4}}^p v_{n\ldots n i_k \ldots i_5i_2i_1p}\Big )\right\} \\&= \frac{(m-2)\sigma (n, \ldots n ,i_k ,\ldots , i_3)}{m(m-1)}\left\{ 2\sum _{q=1}^2\left( \sum _{l=3}^{k}\Gamma _{i_li_q}^p v_{n\ldots n i_k\ldots {\hat{i_l}} \dots {\hat{i_q}}i_1p} +\Gamma _{i_3i_{q}}^p v_{n\dots n i_k \ldots i_4{\hat{i_q}}i_1p} \right) \right. \\&\quad + \left. (m-3)\left( \sum _{l=4}^{k} \Gamma _{i_li_3}^p v_{n\ldots n i_k\ldots {\hat{i_l}} \ldots i_4i_2i_1p}+2\Gamma _{i_3i_4}^p v_{n\ldots n i_k \ldots i_5i_2i_1p}\right) \right\} \\&\quad +\frac{2(k-1)}{m(m-1)} \Gamma _{i_2i_{1}}^p v_{n\dots n i_k\dots i_3p}\\&= \frac{(m-3)\sigma (n, \dots n ,i_k ,\ldots , i_4)}{m(m-1)}\left\{ 2\sum _{q=1}^3\left( \sum _{l=4}^{k}\Gamma _{i_li_q}^p v_{n\dots n i_k\ldots {\hat{i_l}} \ldots {\hat{i_q}}i_1p}\right) \right. \\&\quad \left. +\, 4\sum _{q=1}^3\left( \Gamma _{i_4i_{q}}^p v_{n\ldots n i_k \dots {\hat{i_q}}i_1p} \right) \right. \\&\quad + \left. (m-4)\left( \sum _{l=5}^{k} \Gamma _{i_li_4}^p v_{n\dots n i_k\dots {\hat{i_l}} \dots i_1p}+3\Gamma _{i_5i_4}^p v_{n\dots n i_k \dots i_1p}\right) \right\} \\&\quad +\frac{2(k-1)}{m(m-1)}\sum _{q,r =1, q\ne r}^3\Gamma _{i_qi_{r}}^p v_{n\ldots n i_k\ldots {\hat{i_q}}{\hat{i_r}} i_1p}\\&\text{ Repeating } \text{ this } \text{ expansion } \text{ for } \text{(k-2) } \text{ times } \text{ more } \text{ to } \text{ get } \end{aligned}$$
$$\begin{aligned} J^1_{k}&= \frac{(m-k+1)\sigma (n, \ldots , n ,i_{k})}{m(m-1)}\left\{ 2\sum _{q=1}^{k-1}\Gamma _{i_ki_q}^p v_{n\dots n i_{k-1}\ldots {\hat{i_l}} \ldots i_1p} +2(k-2)\sum _{q=1}^{k-1}\Gamma _{i_ki_q}^p v_{n\ldots n i_{k-1}\dots {\hat{i_l}} \dots i_1p}\right. \\&\quad \left. + (m-k)(k-1)\Gamma _{ni_k}^p v_{n\ldots n i_{k-1} \ldots i_1p}\right\} +\frac{2(k-1)}{m(m-1)}\sum _{q,r =1, q\ne r}^{k-1}\Gamma _{i_qi_{r}}^p v_{n\ldots n i_k\ldots {\hat{i_q}}{\hat{i_r}} i_1p}\\&= \frac{(m-k+1)\sigma (n, \ldots , n ,i_{k})}{m(m-1)}\left\{ 2(k-1)\sum _{q=1}^{k-1}\Gamma _{i_ki_q}^p v_{n\dots n i_{k-1}\ldots {\hat{i_q}} \ldots i_1p}\right. \\&\quad \left. + (m-k)(k-1)\Gamma _{ni_k}^p v_{n\ldots n i_{k-1} \ldots i_1p}\right\} +\frac{2(k-1)}{m(m-1)}\sum _{q,r =1, q\ne r}^{k-1}\Gamma _{i_qi_{r}}^p v_{n\ldots n i_k\ldots {\hat{i_q}}{\hat{i_r}} i_1p}\\&= \frac{2(k-1)}{m(m-1)}\sum _{q,r =1, q\ne r}^k\Gamma _{i_qi_{r}}^p v_{n\ldots n i_k\ldots {\hat{i_q}}\ldots {\hat{i_r}}\ldots i_1p}+\frac{2(k-1)(m-k)}{m(m-1)}\sum _{q=1}^k\Gamma _{ni_q}^p v_{n\dots n i_k\ldots {\hat{i_q}} \ldots i_1p} \end{aligned}$$

After putting the values of \( J, J^1_k\) and \(J_k^2\) in \(\mathrm {d}v\), we get

$$\begin{aligned} (\mathrm {d}v)_{n\dots n i_k\ldots i_1}&=\frac{(m-k)}{m}\frac{\partial v_{n\dots ni_k\ldots i_1}}{\partial x^n} -\frac{2(m-k)}{m} \sum _{l=1}^k\Gamma _{i_l n}^p v_{n\dots n i_k\ldots {\hat{i_l}}\dots i_1 p}\\&+\frac{1}{m}\sum _{l=1}^k \frac{\partial v_{n\dots ni_k\dots {\hat{i_l}}\ldots i_1}}{\partial x^{i_l}}-\frac{2}{m} \sum _{l,q=1, l \ne q}^k\Gamma _{i_l i_q}^p v_{n\ldots n i_k\ldots {\hat{i_l}}\ldots {\hat{i_q}}\ldots i_1 p}. \end{aligned}$$

\(\square \)

Proof of Estimate (11)

Let \(^{t}L= \frac{\Phi _{\xi }\partial _{\xi }}{i\lambda |\Phi _\xi |^{2}}\). Then as already noted

$$\begin{aligned} ^{t}L^{N}(e^{i\lambda \Phi (x,\xi )})=e^{i\lambda \Phi (x,\xi )}. \end{aligned}$$

Consider,

$$\begin{aligned}&\left| \iint _{|x-y|>\delta /C_{0}} (^{t}L^{N}(e^{i\lambda \Phi (y,x,\xi ,\eta )}))\tilde{{\tilde{a}}}_N(x,\xi )f_{i_{1}\ldots i_{m}}(z) {\tilde{b}}^{i_{1}}(x,\xi )\ldots {\tilde{b}}^{i_{m}}(x,\xi )dxd\xi \right| \\&\quad \le \left| \iint _{|x-y|>\delta /C_{0}} e^{i\lambda \Phi (y,x,\xi ,\eta )}L^{N}(\tilde{{\tilde{a}}}_N(x,\xi )f_{i_{1}....i_{m}} (z){\tilde{b}}^{i_{1}}(x,\xi )\ldots {\tilde{b}}^{i_{m}}(x,\xi ))dxd\xi \right| \\&\quad \quad + N\int _{|x-y|>\delta /C_{0}}e^{-\lambda \delta ^{2}/2} \left| f_{i_{1}\ldots i_{m}}(x)B^{i_{1}\ldots i_{m}}(x,\xi _{bdry})\right| dx. \end{aligned}$$

Using the fact that, f is compactly supported and using (7), we get (11). \(\square \)

Proof of the Estimate (13)

Consider

$$\begin{aligned} \left| \iint _{|x-y|<\delta /C_{0}} \left( e^{i\lambda \Phi (y,x,\xi ,\eta )}\right) \tilde{{\tilde{a}}}_N(x,\xi )f_{i_{1}\ldots i_{m}}(z){\tilde{b}}^{i_{1}}(x,\xi )\ldots {\tilde{b}}^{i_{m}}(x,\xi )dxd\xi \right| . \end{aligned}$$

Rewrite the above as :

$$\begin{aligned}&\left| \int _{|x-y|<\delta /C_{0}}(e^{i\lambda \Phi (y,x,\xi _{c},\eta )})(e^{-i\lambda \Phi (y,x,\xi _{c},\eta )})\right. \\&\quad \left. \int _{|\xi -\eta |<\delta /C_{0}} (e^{i\lambda \Phi (y,x,\xi ,\eta )})\tilde{{\tilde{a}}}_N(x,\xi )f_{i_{1}\ldots i_{m}}(z){\tilde{b}}^{i_{1}}(x,\xi )\ldots {\tilde{b}}^{i_{m}}(x,\xi )dxd\xi \right| \end{aligned}$$

Using [18, Remark 2.10], we get

$$\begin{aligned}&\left| \int _{|x-y|<\delta /C_{0}} (e^{i\lambda \Phi (y,x,\xi _{c},\eta )})f_{i_{1}\ldots i_{m}}(x)\right. \\&\quad \left. \sum _{0\le k \le \lambda /C}C_{n}\frac{1}{k!}\lambda ^{-n/2-k}\left( \frac{\Delta }{2}\right) ^k \big (\tilde{{\tilde{a}}}_N(x,\xi _{c}){\tilde{b}}^{i_{1}}(x,\xi _{c}) \ldots {\tilde{b}}^{i_{m}}(x,\xi _{c})\big )\right. \\&\quad \left. +R(x,y,\eta ,\lambda )dx \right| \end{aligned}$$

Lemma 8

$$\begin{aligned} \sum _{0\le k \le \lambda /C}C_{n}\frac{1}{k!}\lambda ^{-n/2-k}\left( \frac{\Delta }{2}\right) ^k \big (\tilde{{\tilde{a}}}_N(x,\xi _{c}){\tilde{b}}^{i_{1}}(x,\xi _{c}) \cdots {\tilde{b}}^{i_{m}}(x,\xi _{c})\big ) \end{aligned}$$

is a formal analytic symbol.

Proof

Let,

$$\begin{aligned} a_{k} = \frac{1}{k!}\left( \frac{\Delta }{2}\right) ^k \big (\tilde{{\tilde{a}}}_N(x,\xi _{c}){\tilde{b}}^{i_{1}}(x,\xi _{c}) \cdots {\tilde{b}}^{i_{m}}(x,\xi _{c})\big ) \end{aligned}$$

Then from the Cauchy integral formula [18, Exercise 2.4],

$$\begin{aligned} |a_{k}|&\le C_{n}(k+1)^{n/2}(k-1)! 2^{k}\sup _{B_1(\xi _c)}\big (\tilde{{\tilde{a}}}_N(x,\xi ){\tilde{b}}^{i_{1}} (x,\xi )\ldots {\tilde{b}}^{i_{m}}(x,\xi )\big )\\&\le C1_{n}(k+1)^{n/2}(k-1)! 2^{k}\\&\le C2_{n}(k+1)^{n/2}e^{2-k}(k-1)^{k-1/2} 2^{k} \text { (Using Stirling's approximation)}\\&\le C2_{n}\bigg (\frac{2}{e}\bigg )^{k+1}(k+1)^{n/2+k}\\&\le \tilde{C_{n}}^{k+n/2}(k+n/2)^{n/2+k}. \end{aligned}$$

Hence,

$$\begin{aligned}&\sum _{0\le k \le \lambda /C}C_{n}\frac{1}{k!}\lambda ^{-n/2-k}\left( \frac{\Delta }{2}\right) ^k \big (\tilde{{\tilde{a}}}_N(x,\xi _{c}){\tilde{b}}^{i_{1}}(x,\xi _{c}) \ldots {\tilde{b}}^{i_{m}}(x,\xi _{c})\big )\\&\qquad = \sum _{0\le k \le \lambda /C}\lambda ^{-n/2-k}a_{k+n/2} \end{aligned}$$

is a formal analytic symbol \({{B}}^{i_{1}\ldots i_{m}}(x,y,\eta ;\lambda )\) by [18, Exercise 1.1]. \(\square \)

Hence,

$$\begin{aligned}&\int _{|x-y|<\delta /C_{0}} (e^{i\lambda \Phi (y,x,\xi ,\eta )})\tilde{{\tilde{a}}}_{N}(x,\xi )f_{i_{1}\ldots i_{m}}(z){\tilde{b}}^{i_{1}}(x,\xi )\ldots {\tilde{b}}^{i_{m}}(x,\xi )dxd\xi \\&\quad = \int _{|x-y|<\delta /C_{0}} (e^{i\lambda \Phi (y,x,\xi _{c},\eta )})f_{i_{1}\dots i_{m}}(x){{B}}^{i_{1}\ldots i_{m}}(x,y,\eta ;\lambda )dx\\&\qquad + \int _{|x-y|<\delta /C_{0}} (e^{i\lambda \Phi (y,x,\xi _{c},\eta )})f_{i_{1}\dots i_{m}}(x)R(x,y,\eta ;\lambda )dx . \end{aligned}$$

But,

$$\begin{aligned} \left| \int _{|x-y|<\delta /C_{0}} (e^{i\lambda \Phi (y,x,\xi _{c},\eta )})f_{i_{1}\dots i_{m}}(x)R(x,y,\eta ;\lambda )dx\right| ={\mathcal {O}}(e^{-\lambda /c}), \end{aligned}$$

since,

$$\begin{aligned} |R(x,y,\eta ;\lambda )|\le |\Omega |/C e^{-\lambda /c}, \end{aligned}$$

see [18, Remark 2.10]. So, this along with (10) and (11), gives us:

$$\begin{aligned} \left| \int _{|x-y|<\delta /C_{0}} (e^{i\lambda \Phi (y,x,\xi _{c},\eta )})f_{i_{1}\dots i_{m}}(x){{B}}^{i_{1}\ldots i_{m}}(x,y,\eta ;\lambda )dx\right| = {\mathcal {O}}(e^{-\lambda /c}). \end{aligned}$$

\(\square \)

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Abhishek, A., Mishra, R.K. Support Theorems and an Injectivity Result for Integral Moments of a Symmetric m-Tensor Field. J Fourier Anal Appl 25, 1487–1512 (2019). https://doi.org/10.1007/s00041-018-09649-7

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  • DOI: https://doi.org/10.1007/s00041-018-09649-7

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