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In MTW "Gravitation", the projection tensor is defined as

$$\boldsymbol{P} = \boldsymbol{g} + \boldsymbol{u}\otimes\boldsymbol{u}$$

And one exercise asks to prove that a tangent vector $\boldsymbol{B}$ projected by $\boldsymbol{P}$ is orthogonal to $\boldsymbol{u}$. They suggest using an orthonormal frame and that $\boldsymbol{B} = B^{\alpha}\boldsymbol{e}_{\alpha}$ and $\boldsymbol{e}_0 = \boldsymbol{u}$. Then, $\boldsymbol{P}\cdot\boldsymbol{B} = B^{j}\boldsymbol{e}_j$, so purely spatial.

My question is that I don't know how to interpret something like $\boldsymbol{g}\cdot\boldsymbol{B}$ , the product of a 2-tensor with a vector. It is not a contraction. It seems also weird to me the $\boldsymbol{u}\otimes\boldsymbol{u}\cdot\boldsymbol{B}$.

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  • $\begingroup$ It is not a contraction. Yes, it is. $\endgroup$ – G. Smith Sep 2 at 0:56
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The notation $\mathbf{P}\cdot\mathbf{u}$ simply means to interpret $\mathbf{P}$ as a linear transformation and to let it act on $\mathbf{u}$, or equivalently to take the dot product on the last index of $\mathbf{P}$:

$$(\mathbf{P}\cdot\mathbf{u})^\mu = P^\mu{}_\nu u^\nu = P^{\mu\rho} g_{\rho\nu} u^\nu.$$

So addressing your examples, we first have $(\mathbf{g}\cdot\mathbf{B})^\mu = g^{\mu\rho} g_{\rho\nu} B^\nu = \delta^\mu{}_\nu B^\nu = B^\mu$. That is, when converted to a $(1,1)$ tensor, the metric is just the identity transformation. And the second, more interesting term is $((\mathbf{u}\otimes\mathbf{u})\cdot\mathbf{B})^\mu = (\mathbf{u}\otimes\mathbf{u})^\mu{}_\nu B^\nu = u^\mu u_\nu B^\nu$.

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