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Sancol.
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Proof of this relation depends on what approach is your textbook used to learn tensor calculus, for example, If you already defined the action of $g_{\mu \nu}$ for lowering indexes, now you can write $$g_{\mu\nu}g^{\nu\alpha}=g_{\mu\nu}e^\nu \cdot e^\alpha = e_\mu \cdot e^\alpha = \delta^\alpha_\mu$$

This proof is in Core Principles of Special and General Relativity by Luscombe:

Proof of this relation depends on what approach is your textbook used to learn tensor calculus, for example, If you already defined the action of $g_{\mu \nu}$ for lowering indexes, now you can write $$g_{\mu\nu}g^{\nu\alpha}=g_{\mu\nu}e^\nu \cdot e^\alpha = e_\mu \cdot e^\alpha = \delta^\alpha_\mu$$

Proof of this relation depends on what approach is your textbook used to learn tensor calculus, for example, If you already defined the action of $g_{\mu \nu}$ for lowering indexes, now you can write $$g_{\mu\nu}g^{\nu\alpha}=g_{\mu\nu}e^\nu \cdot e^\alpha = e_\mu \cdot e^\alpha = \delta^\alpha_\mu$$

This proof is in Core Principles of Special and General Relativity by Luscombe:

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Sancol.
  • 964
  • 1
  • 5
  • 20

Proof of this relation depends on what approach is your textbook used to learn tensor calculus, for example, If you already defined the action of $g_{\mu \nu}$ for lowering indexes, now you can write $$g_{\mu\nu}g^{\nu\alpha}=g_{\mu\nu}e^\nu \cdot e^\alpha = e_\mu \cdot e^\alpha = \delta^\alpha_\mu$$