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One of the definition of tensor is that it should be invariant under transformation. Thus a tensor of order zero is a scalar should be invariant under transformation. For a tensor component $v^{\mu}$ the inner product $v^{\mu}v_{\mu}$ is a tensor of rank zero and should be invariant under Lorentz transformation.But, $$v'^{\mu}v'_{\mu}={\Lambda_{\mu}}^{\alpha}{\Lambda^{\mu}}_{\beta}v^{\beta}v_{\alpha}$$
${\Lambda_{\mu}}^{\alpha}{\Lambda^{\mu}}_{\beta} $ not equal to ${\delta_{\beta}}^{\alpha}$, which implies $v'^{\mu}v'_{\mu}$ not equal to $v^{\mu}v_{\mu}$. This shows that $v^{\mu}v_{\mu}$ is not invariant under transformation(not a tensor). Can anybody suggest where am I wrong?

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  • $\begingroup$ I'm sorry. Its typing mistake. $\endgroup$ – walber97 Feb 26 at 17:54
  • $\begingroup$ I want to know wheather the inner product of two tensors which contract to a zero order tensor is invariant under transformation. $\endgroup$ – walber97 Feb 26 at 17:57
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    $\begingroup$ Yes, it is. To see how, first you have to be careful about index placement. You cannot just write $\Lambda^\alpha_\mu$ because ${\Lambda^\alpha}_\mu$ is not the same thing as ${\Lambda_\mu}^\alpha$. Second, you need to write down the defining relation for $\Lambda$ to be a Lorentz transformation. $\endgroup$ – G. Smith Feb 26 at 18:02
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    $\begingroup$ OK. Do you know what relation $\Lambda$ has to satisfy to be a Lorentz transformation? $\endgroup$ – G. Smith Feb 27 at 3:00
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    $\begingroup$ Right. Work that out in index notation and you should find that ${\Lambda_{\mu}}^{\alpha}{\Lambda^{\mu}}_{\beta} $ *is* equal to ${\delta_{\beta}}^{\alpha}$ because the mixed-index ${\eta^\mu}_\nu$ is ${\delta^\mu}_\nu$. $\endgroup$ – G. Smith Feb 27 at 3:12

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