# What does the Pythagorean Theorem really prove?

The question adds up to the following questions and I hope it’s not considered duplicate:

Where does the Pythagorean theorem "fit" within modern mathematics?

Why does the Pythagorean Theorem have its simple form only in Euclidean geometry?

Let $(E, d)$ be a metric space.

If $d$ is the Euclidean metric, then the Pythagorean Theorem is essentially the definition of $d$, or a trivial corollary.

If $d$ is not the Euclidean metric, then the Pythagorean Theorem does not hold (is this true?).

So: what is the underlying truth of the Pythagorean Theorem, in modern mathematical terms?

Note: since all the proofs are implicitly based on the invariance of the metric with respect to translations and rotations, I thought that the underlying “theorem” was that the Euclidean metric is the only one whose group of isometries is the Euclidean Group. This is in line with some of the answers to the questions, but this is not true (for a counterexample see

I would be satisfied with an answer of the type: the Euclidean metric is the only metric that comes from an inner product (and therefore the set $E$ can be made a vector space), that is invariant with respect to rotations, translations and reflections, [+ more conditions].

Or please give a reference, I feel that this is pretty basic but I can't figure it out! Thanks.

• The Pythagorean theorem can be understood as a theorem about norms coming from inner products. In this guise it has a useful infinite-dimensional version, namely Parseval's identity. – Qiaochu Yuan Mar 7 '17 at 23:15
• When we define the Euclidean metric $d(x,y) = \| x - y \|_2$, it's not obvious that this is a reasonable model of distance in the real world. But Hilbert's axioms for Euclidean geometry do seem like a reasonable model of the world, and from these axioms the distance formula can be derived. – littleO Mar 7 '17 at 23:18
• You can't even state the question of the Pythagorean theorem in an arbitrary metric space, because angles are not defined in an arbitrary metric space. So your second question -- "If $d$ is not the Euclidean metric, then the Pythagorean Theorem does not hold" -- doesn't make sense. For the easiest example, let $E$ be any set and let $d$ be the discrete metric. You can't even ask whether, in this geometry, the sum of the squares of the lengths of the legs of a right triangle is the square of the hypotenuse, because you can't even talk about "right triangles." – symplectomorphic Mar 8 '17 at 0:42

The basic reference for the significance of the Pythagorian theorem is the illuminating article by Givental:

• Love it!!! Allow me to quote what I was looking for: "Among all Riemannian metrics of constant curvature only the Euclidean one admits nontrivial conformal isometries." And this is exactly what triggered my question, having to teach the Theorem to kids: "Yet it [the typical geometric proof] pictures the whole issue as a cut-and-paste puzzle and leaves us with a feeling of disproportion: one of the most fundamental facts of nature is due to an ingenious tiling trick." Definetly a MUST READ, thanks. – Giannakos Mar 8 '17 at 16:51

Let $V$ be a real or complex vector space equipped with an inner product $\langle -, - \rangle$, and let $\| - \|$ be the corresponding norm. One version of the Pythagorean theorem states that if $v_1, \dots v_n$ are mutually orthogonal vectors in $V$ (meaning that $\langle v_i, v_j \rangle = 0$ for $i \neq j$), then

$$\left\| \sum v_i \right\|^2 = \sum_i \| v_i \|^2.$$

This theorem has content even in the case that $V = \mathbb{R}^d$ and $\langle -, - \rangle$ is the dot product, the point being that $v_1, \dots v_n$ are arbitrary and not necessarily aligned with the coordinate axes. It has various uses in other more exotic cases; for example, for random variables of mean zero equipped with the covariance inner product, this statement is linearity of variance. And of course there is the version with infinitely many $v_i$ for use in Hilbert spaces, e.g. to prove Parseval's identity.

From this coordinate-free point of view you can't even define rotations and reflections without a choice of inner product, so it's unclear how to interpret your last question.

• Thank you for the answer, this is a very nice generalization. From this and the comments above, I extract something "obvious" that I had not noted. I was thinking in terms of $(E,d)$, $E$ a being merely a set. But now I notice that, for the Pythagorean Theorem to make sense, $E$ has to be a vector space with inner product (otherwise there is no right triangle definition). Therefore, it is just a particular case of your general version, which holds for all inner product spaces. This is very satisfactory. [continues] – Giannakos Mar 8 '17 at 14:37
• There’s still one aspect that disturbs me. Let's think of the simple geometric high school proofs of the Theorem. Basically, one chops a square into triangles and rearranges them around. This doesn't use any notion of scalar product (altohugh it must be hiden somewhere). As far as I can see, it just uses the Euclidean group isometries, the fact that the lengths of the sides remain the same when I move the triangles. But the isometries alone don’t fix the Euclidean metric! So what am I missing? What is the other ingredient in the geometric proofs besides moving triangles around? – Giannakos Mar 8 '17 at 14:37
• @Giannakos: that's a good question. I have two hypotheses although I haven't thought about it in detail: first, the standard dissection proofs start with a square, and to define squares you need to know what right angles are in addition to lengths, which is close to knowing an inner product. Second, the dissection proofs involve areas, and it's not at all clear how to define areas given only, say, a norm on $\mathbb{R}^2$ as opposed to an inner product. (This is related to the first point: the most basic area to define is the area of a rectangle, which we need right angles to define.) – Qiaochu Yuan Mar 8 '17 at 22:30
• Yes! The scalar product is in the chopping, "bisecting the square", it seems so obvious now, thanks! I'll think of it more deeply. – Giannakos Mar 9 '17 at 20:38

I rather love the question you linked concerning the behavior of triangles in non-Euclidean geometry. This may not be exactly what you are looking for, but I think it really forms the basis for why the Pythagorean theorem is special. The other answer gives a cool algebraic answer in some sense, so here is a geometric one.

The basic idea is this:

Euclidean space is the "sweet point" where "straight lines" (geodesics) satisfy the Pythagorean theorem. This is because it is the "transition point" between positively and negatively curved space (i.e. where there is no curvature).

Suppose we have a point $x$ and start two geodesics moving at constant unit speed from each other with angle $\theta$. If you look in Villani, Optimal transport, old and new, eq 14.1 (also see this question), for 2-manifolds, we get the distance $d(t)$ between the two moving points at time $t$ to be: $$d(t) = t\sqrt{2(1-\cos(\theta))} \left[ 1 - \frac{\kappa(x)\,\cos^2(\theta/2)}{6}t^2 + O(t^4) \right]$$ where $\kappa(x)$ is the Gaussian curvature at $x$.

Let's construct a right triangle, with two sides of equal length $t$ and measure the hypotenuse as the distance between them. Assume $t<<1$. Take $\theta=\pi/2$. Then: $$d(t) = t\sqrt{2}\left[ 1 - \frac{\kappa(x)\,t^2}{8} \right]$$ is the hypotenuse length (i.e. $c=d(t)$).

Recall that the other two sides of our triangle are of length $a=b=t$. Notice that if $\kappa=0$, we get: $$a^2+ b^2 = t^2+t^2 = 2t^2=d(t)^2 = c^2$$ so on an uncurved manifold, Pythagorean theorem is correct!

Note that everything depends on the curvature of the space. If $\kappa(x)<0$, then $a^2+b^2 < c^2$; if $\kappa(x)>0$, then $a^2+b^2 > c^2$. Indeed, the degree to which Pythagoras is accurate decreases as space becomes more curved.

So, my answer to what is the underlying truth of the Pythagorean Theorem, in modern mathematical terms, is that I would consider the Pythagorean theorem to be the special case result of geodesic distances in uncurved space.

This holds somewhat more generally for Riemannian manifolds: negative Ricci scalar curvature causes geodesics to diverge, while positive does the opposite. Hence, if we construct "triangles" by treating "straight lines" to be geodesics and measure the distance between them as they grow, in the former case, the distance will "overshoot" the Pythagorean theorem; in the latter case, it will undershoot. (There's a cool little graph of that in here, albeit in a probabilistic sense). Only in the flat Euclidean case (at least locally), will they be exactly equal all the time.

Maybe some people will also want to note that, in geodesic normal coordinates, the metric is locally euclidean (i.e. $g_{ij} \approx \delta_{ij}$, so the Pythagorean theorem will hold infinitesimally (since the square geodesic distance will be $s^2=g_{ij}x^ix^j \approx \delta_{ij}x^ix^j=(x^1)^2 + (x^2)^2$ in 2D).