# Projection tensor in General Relativity

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 $$\boldsymbol{u}\otimes\boldsymbol{u}\cdot\boldsymbol{B}$$.

• It is not a contraction. Yes, it is. – G. Smith Sep 2 at 0:56

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 $$((\mathbf{u}\otimes\mathbf{u})\cdot\mathbf{B})^\mu = (\mathbf{u}\otimes\mathbf{u})^\mu{}_\nu B^\nu = u^\mu u_\nu B^\nu$$.