I'm attempting to prove that, if $U=(U^i_j)_{n \times n}$ is the matrix associated to an orthogonal transformation and $g=(g_{ij})_{n \times n}$ is the metric tensor, then

$$U^{T} g U=g \tag{1}$$

My attempt

Applying the condition that an orthogonal transformation must preserve the scalar product, $\big(U(\boldsymbol{x}), U(\boldsymbol{y})\big) = (\boldsymbol{x}, \boldsymbol{y})$, I have found that that

$$ \sum ^n _{i=1} \sum ^n _{j=1} \sum ^n _{k=1} \sum ^n _{l=1} g_{i j} U_{k}^{i} U_{l}^{j} x^{k} y^{l}= \sum ^n _{i=1}\sum ^n _{j=1} g_{i j} x^{i} y^{j} \tag{2}$$

Using the Einstein summation convention and changing in the RHS the indexes $i \rightarrow k, j \rightarrow l$

$$g_{i j} U_{k}^{i} U_{l}^{j} x^{k} y^{l} = g_{kl} x^{k} y^{l} \tag{3}$$

Since the terms $x^{k} y^{l}$ appears at both sides, if we could identify $g_{i j} U_{k}^{i} U_{l}^{j}=g_{k l}$, this would correspond to the matrix product $U^{T} g U=g$, and the proof would be complete.

But I'm stuck here, because I don't see how we can match parts of different sums... Could someone tell me how could it be justified from $(3)$ that $g_{i j} U_{k}^{i} U_{l}^{j}=g_{k l}$ (if it is actually possible to do this)?

  • 2
    $\begingroup$ This doesn't hold for just some $\boldsymbol x, \boldsymbol y$. It holds for every $\boldsymbol x, \boldsymbol y$. $\endgroup$ Jan 3, 2021 at 20:46

2 Answers 2


@PaulSinclair is right. Suppose $x^k$ ($y^l$) is $0$ except for being $1$ when $k=m$ ($l=n$) in our coordinates, so $g_{ij}U^i_mU^j_n=g_{mn}$.

  • $\begingroup$ Okay, but with this reasoning you only show that it is true for the particular case that $\boldsymbol{x}$ and $\boldsymbol{y}$ are vectors of the canonical base, doesn't you? $\boldsymbol{x}$ or $\boldsymbol{y} = (1,0,...,0), (0,1,...,0)$, etc. $\endgroup$
    – Invenietis
    Jan 4, 2021 at 10:03
  • 1
    $\begingroup$ @user206148 The value of $g_{ij}U^i_mU^j_n$ doesn't depend on what you contract it with. The only way for $g_{ij}U^i_kU^j_lx^ky^l=g_{kl}x^ky^l$ to be true for all choices of $x,\,y$, including mine, is to have $g_{ij}U^i_kU^j_l=g_{kl}$. $\endgroup$
    – J.G.
    Jan 4, 2021 at 10:46

As already said, the key is that the equality must be satisfied for any values of $\boldsymbol{x}=(x^1,...,x^n)$ and $\boldsymbol{y}=(y^1,...,y^n)$. Also notice that in the LHS, once the sum is performed over $i$ and $j$, there is a term that only depends on $k$ and $l$. This is clear by reordering equation $(3)$ as follows:

$$ \sum ^n _{k=1} \sum ^n _{l=1} x^{k} y^{l} \bigg(\sum ^n _{i=1} \sum ^n _{j=1} g_{i j} U_{k}^{i} U_{l}^{j} \bigg) = \sum ^n _{k=1}\sum ^n _{l=1} x^{k} y^{l}g_{kl} $$

$$ \sum ^n _{k=1} \sum ^n _{l=1} x^{k} y^{l} h_{kl} = \sum ^n _{k=1}\sum ^n _{l=1} x^{k} y^{l}g_{kl} $$

So $h_{kl}=g_{i j} U_{k}^{i} U_{l}^{j}=g_{kl}$.


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