Contracting tensors with the metric $g$

Question

In Riemannian geometry I often see some tensor being contracted with the metric $g$. Now I'm not entirely sure what is being meant by contracting using the metric and I cannot find a reasonable explanation of this. If I have a $(1,2)$-tensor $$A=A^j_{ik}dx^i \otimes \partial_j\otimes dx^k,$$ can I contract this using the metric in some way?

Answer 1

Suppose $V_1,\dots, V_k,W$ ($k\geq 2$) are (finite-dimensional) vector spaces over a field $\Bbb{F}$, fix two distinct indices $i,j\in\{1,\dots, k\}$, and suppose we have a bilinear map $\beta:V_i\times V_j\to W$. Then, there is a unique linear map \begin{align} C_{\beta,i,j}:V_1\otimes\cdots\otimes V_k\to W\otimes V_1\otimes \cdots\widehat{V_i}\otimes\cdots\otimes\widehat{V_j}\otimes\cdots\otimes V_k \end{align} (where the hats means omission) such that the action on pure tensors is \begin{align} v_1\otimes\cdots\otimes v_k\mapsto\beta(v_i,v_j)\otimes v_1\otimes \cdots\widehat{v_i}\otimes\cdots\otimes\widehat{v_j}\otimes\cdots\otimes v_k. \end{align}

This map $C_{\beta,i,j}$ is called the $(i,j)$ trace/contraction on $V_1\otimes\cdots\otimes V_k$ relative to $\beta$. This is a very general definition, so let us now extract several special cases

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Example 1: $W=\Bbb{F}$

In this case, we have $C_{\beta,i,j}:V_1\otimes\cdots\otimes V_k\to V_1\otimes \cdots\widehat{V_i}\otimes\cdots\otimes\widehat{V_j}\otimes\cdots\otimes V_k $, and its action on pure tensors is \begin{align} v_1\otimes\cdots\otimes v_k\mapsto\beta(v_i,v_j)\cdot v_1\otimes \cdots\widehat{v_i}\otimes\cdots\otimes\widehat{v_j}\otimes\cdots\otimes v_k. \end{align} i.e in this special case $\beta(v_i,v_j)$ is just a number so we can simply scalar multiply it with the tensor $v_1\otimes\cdots\otimes\widehat{v_i}\cdots\otimes\widehat{v_j}\otimes\cdots\otimes v_k$.

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Example 2: $k=2$

So, here we only have two vector spaces $V_1,V_2$ and a bilinear map $\beta:V_1\times V_2\to W$. In this case, the trace on $V_1\otimes V_2$ relative to $\beta$ is simply the unique linear map $V_1\otimes V_2\to W$ such that the action on pure tensors is \begin{align} v_1\otimes v_2\to \beta(v_1,v_2). \end{align} In other words, it is the unique linear map given to us by the universal properties of tensor products, where the bilinear map descends to a linear map out of the tensor product.

Example 2.1: $k=2$, $V_1,V_2$ are dual spaces, $W=\Bbb{F}$, $\beta$ is evaluation

This is perhaps the most important example of all. Let us call $V_2=V$ and suppose $V_1=V^*$. Then, there’s the super important bilinear map, given by evaluation $\beta=\text{ev}:V^*\times V\to\Bbb{F}$, $(\phi,v)\mapsto \phi(v)$. This gives us a unique linear map $V^*\otimes V\to\Bbb{F}$, and after using the canonical isomorphism $V^*\otimes V\cong \text{End}(V)$, this literally becomes the trace functional $\text{tr}:\text{End}(V)\to\Bbb{F}$ that we all learn in linear algebra (take the matrix of the operator and sum the diagonals).

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Example 3: k=3.

This is relevant to what you’re asking. Fix a pseudo-inner product space $(V,g)$. Now, define $(V_1,V_2,V_3)=(V^*,V,V^*)$. So, we now have a triple tensor product space $V^*\otimes V\otimes V^*$. There are many types of contractions we can take

• Suppose we want to take the $(1,2)$-contraction. Then, we can use the above evaluation map $\beta_{1,2}=\text{ev}:V^*\times V\to\Bbb{R}$. The net result is the contraction map $C_{\text{ev},1,2}:V^*\otimes V\otimes V^*\to V^*$ given by $\phi\otimes v\otimes \psi\mapsto \text{ev}(\phi,v)\cdot\psi=\phi(v)\cdot \psi$. So, this contraction takes a $(1,2)$ tensor (an element of $V^*\otimes V\otimes V^*$) and produces a $(0,1)$ tensor (an element of $V^*$).
• We can similarly take the $(2,3)$-contraction $C_{\text{ev},2,3}:V^*\otimes V\otimes V^*\to V^*$, whose action on pure tensors is $\phi\otimes v\otimes \psi\mapsto \psi(v)\cdot \phi$. Note that although the $(1,2)$ and $(2,3)$ contractions have the same domain and target space, these are different maps (they differ by a flip of the two $V^*$’s on the domain).

These are the most obvious contractions, and so far we haven’t used the metric $g$ yet. Now, we can ask whether we can define a $(1,3)$-contraction. For this, we’re going to need a bilinear map $V^*\times V^*\to W$ for some vector space $W$. There is no natural such map lying around in general. However, because we have the metric $g:V\times V\to\Bbb{R}$, we can first of all set $W=\Bbb{R}$, and then use the musical isomorphism $g^{\flat}:V\to V^*$, $x\mapsto g(x,\cdot)$, to define a bilinear map $\beta:=\widetilde{g}:V^*\times V^*\to\Bbb{R}$. This is the ‘inverse metric’, with components $g^{ab}$ (see here for more details).

• With this in mind, we can define the $(1,3)$-contraction $C_{\widetilde{g},1,3}:V^*\otimes V\otimes V^*\to V$ whose action on pure tensors is $\phi\otimes v\otimes \psi\mapsto \widetilde{g}(\phi,\psi)\cdot v$.

So, in particular, if you fix a basis $\{e_1,\dots, e_n\}$ for $V$, and let $\{\epsilon^1,\dots,\epsilon^n\}$ be the dual basis, and you have a tensor $A\in V^*\otimes V\otimes V^*$ whose components are given as \begin{align} A=A_{i\,\,k}^{\,\,j}\cdot \epsilon^i\otimes e_j\otimes \epsilon^k, \end{align} then applying this $(1,3)$-contraction (which highly depends on the metric $g$, so it’s also called the metric contraction/trace over the first and third slots) gives \begin{align} C_{\widetilde{g},1,3}(A)&= C_{\widetilde{g},1,3}(A_{i\,\,k}^{\,\,j}\cdot \epsilon^i\otimes e_j\otimes \epsilon^k)\\ &= A_{i\,\,k}^{\,\,j}\cdot C_{\widetilde{g},1,3}(\epsilon^i\otimes e_j\otimes \epsilon^k)\tag{$C_{\widetilde{g},1,3}$ is linear}\\ &= A_{i\,\,k}^{\,\,j}\cdot \widetilde{g}(\epsilon^i,\epsilon^k)\cdot e_j\\ &\equiv A_{i\,\,k}^{\,\,j}\cdot g^{ik}\cdot e_j. \end{align}

So far we’ve been discussing what happens at the level of vector spaces; if you do this at every tangent space of your manifold, you get the corresponding equation at the level of tensor fields: the tensor field $A= A_{i\,\,k}^{\,\,j} \,dx^i\otimes \partial_j\otimes dx^k$ is sent to the vector field $A_{i\,\,k}^{\,\,j} \cdot g^{ik}\,\partial_j$.

A somewhat more common notation for this is $\text{tr}_{g,1,3}$, instead of $C_{\widetilde{g},1,3}$, to mean the trace relative to $g$ over the first and third slots (even though technically speaking, it is the trace relative to the induced map $\widetilde{g}$… but saying all this just makes it more cumbersome, so we don’t).

Answer 2

Think of a $(1,2)$-tensor as a vector-valued bilinear form. Then $A(Y,\cdot,Z) = \sum A^j_{ik}Y^iZ^k\partial_j$. Define a $(0,3)$-tensor by $$B(X,Y,Z)=g\big(X,A(Y,\cdot,Z)\big).$$ In tensor notation, we have $$B_{i\ell k} = \sum g_{j\ell}A^j_{ik}.$$

Answer 3

First, consider only $2$-tensors. A covariant $2$-tensor on a vector space $T$ (e.g., tangent space) is a bilinear function $$B: T\times T \rightarrow \mathbb{R}.$$ It extends uniquely to a linear map $$ B: T\otimes T \rightarrow \mathbb{R}. $$ Therefore, $B$ is a function of contravariant $2$-tensors. Given any contravariant $2$-tensor $A \in T\otimes T$, the contraction of $B$ with $A$ is $B(A)$.

A Riemannian metric is a positive definite covariant $2$-tensor on $T$, $$ g: T\otimes T \rightarrow \mathbb{R}. $$However, since it is non-degenerate, it has an inverse, which is a contravariant $2$-tensor, $$ g^{-1} \in T\otimes T. $$ The contraction of a covariant $2$-tensor $B$ by the metric $g$ means $B(g^{-1})$.

If a basis of $T$ is $(\partial_1,\dots, \partial_n)$ and the dual basis is $(dx^1, \dots, dx^n)$, then we can write $B = b_{ij}\,dx^i\otimes dx^j.$ If $g = g_{ij}\,dx^i\otimes dx^j$, then its inverse is $g^{-1} = g^{ij}\,\partial_i\otimes\partial_j.$ Then the contraction of $B$ by $g$ is $$ b_{ij}g^{ij}. $$

If you have a more complicated tensor $M$ with at least one pair of lower indices, then the tensor can be viewed as a bilinear map $$ M: T\otimes T \rightarrow W,$$ where $W$ is a vector space (that might consist of tensors). Then the contraction of $M$ by $g$ is again $M(g^{-1})$. For specific examples, you can work out what this looks like with respect to a basis of $T$.

Answer 4

Cantraction means to replace the first tensor sign by evaluation $$dx^i\otimes \partial_k \to dx^i(\partial_k) = \delta^i_k$$ yielding $A^i_{ik} dx^k$. The evaluation is identical with the scalar product of the two vectors $$dx^i(\partial_k) = g^{ik}\partial_i \partial_k $$