Methane In $\mathbb{R}^4$ ?

Denote $n+1$ non-zero unit vectors in $\mathbb{R}^n$ by $\{\mathbf{u}_i, i=1,2, \ldots, n+1\}$ . Define the center of gravity of these vectors by $\mathbf{g}=\frac{1}{n+1} \sum_{i=1}^{n+1} \mathbf{u_i}$ If the common cosines between the vectors $\{\mathbf{u}_i\}$ be denoted by $\cos(\theta_n)$ , then for all $i=1,2,\ldots, n+1$ we have $\mathbf{u}_i.\mathbf{g}=\frac{1}{n+1} (1 + n \cos(\theta_n) )$ Since the vectors $\{\mathbf{u}_i, i=1,2,\ldots,n+1\}$ lives in an $n$ -dimensional space, they must be linearly dependent . Thus there exists scalars $\lambda_i$ , not all-zero, such that $\sum_{i=1}^{n+1}\lambda_i \mathbf{u}_i=0$ Taking dot product of the above equation with $\mathbf{g}$ , we conclude that $(1+ n\cos(\theta_n))\bigg(\sum_{i=1}^{n+1}\lambda_i\bigg)=0$ Which implies $\cos(\theta_n)=-\frac{1}{n}$ provided $\sum_{i=1}^{n+1}\lambda_i\neq 0$ , i.e., the vectors are affinely independent, i.e., they enclose a strictly positive volume in $\mathbb{R}^n$ . See my comment below for a proof.

Hence, for $n=4$ , the answer is $-0.25$ and $\lim_{n \to \infty} \theta_n \to \frac{\pi}{2}$ .

My solution is inductive in the number of dimensions $n$ . It is a more general solution than others have given, because it assumes no more than a sequence of nested inner product spaces of increasing dimension.

Suppose that in $n$ -dimensional space we have $n+1$ vectors, $u_0, \dots u_n$ such that $\begin{cases} u_i^2 = 1 \\ u_i\cdot u_j = -k \end{cases}$ for all $0 \leq i \not= j \leq n$ and some value $k = -\cos \theta_n$ . (Note: $u_i\cdot u_j$ is the inner product and $u_i^2 = |u_i|^2 = u_i\cdot u_i$ is the corresponding square norm.)

Now we go to $n+1$ -dimensional space, which contains our original space as a subspace (preserving its inner product); let $z$ be a unit vector in the new space orthogonal to the original space. Define $\begin{cases} v_i = \alpha\:u_i + \beta\:z & 0 \leq i \leq n \\ v_{n+1} = -z\end{cases}$ Here, $\alpha$ and $\beta$ are parameters to determined, in such a way that the $v_i$ obey the same relations as the $u_i$ (see above), but with a different value for $k$ . It is sufficient to require that $\begin{cases} v_i^2 = 1 \\ v_i\cdot v_{n+1} = v_i\cdot v_j = -k'\end{cases}$ for some $0 \leq i \not= j \leq n$ .

The first condition translates into $\alpha^2 + \beta^2 = 1$ ; the second condition gives $(\alpha\:u_i + \beta\:z)\cdot (-z) = (\alpha\:u_i + \beta\:z)\cdot (\alpha\:u_j + \beta\:z) \\ -\beta = -\alpha^2\:k + \beta^2.\\ (1+k)\beta^2 + \beta - k = 0 \\ \beta = -1\ \text{or}\ \frac k{1+k}.$ The first solution would make all vectors $v_i$ equal to $z$ , which all make the same angle of $0^\circ$ with each other; this is not the solution we are looking for. Thus we go with the second solution and have $k' = \beta = \frac k{1+k};\ \ \alpha = \frac 1{1+k}.$

Now we have a way to "upgrade" an $n$ -dimensional solution to an $n+1$ -dimensional solution, and a way to see how $k = -\cos \theta_n$ behaves under this transformation.

• For the basic case $n = 1$ , the only possible non-zero angle between vectors is $180^\circ$ , with $k = 1$ .

• For $n = 2$ we get $k = 1/2$ corresponding to a $120^\circ$ angle.

• For $n = 3$ we get $k = 1/3$ corresponding to a $109.5^\circ$ angle.

• It is not difficult to show that in general $k = 1/n$ . As $n \to \infty$ , $k \to 0$ , so that the angle approaches $90^\circ$ from above.

The solution to the problem, for four dimensions, is $\cos\theta_4 = -k = -\tfrac14$ or $\boxed{0.25}$ .

An elementary proof by finding a recursion on $\theta_n$ .

Let $\theta_n$ denotes the desired angle associated with the $n+1$ unit vectors, $\{\mathbf{x}_1,\cdots,\ \mathbf{x}_{n+1}\}$ living in $\mathbb{R}^n$ , as mentioned in the question. W.l.o.g. we can assume that $\mathbf{x}_1=[1,0,\cdots,0]$ . This forces the other vectors to have the following form: $\mathbf{x}_{k+1}=[\cos\theta_n\ \mathbf{x}_k^{(n-1)}],\quad k=1,2,\cdots,\ n$ Consequently, we have a set of $n$ vectors $\{\mathbf{x}_1^{(n-1)},\cdots,\ \mathbf{x}_{n}^{(n-1)}\}$ , such that $\langle\mathbf{x}_i^{(n-1)}, \mathbf{x}_j^{(n-1)}\rangle=\cos\theta_n-\cos^2\theta_{n},\ \forall i\ne j$ . However, the vectors $\mathbf{x}_i^{(n-1)}$ , thus obtained, are not normalized. By dividing them with their norm (which is $|\sin\theta_n|$ as per the construction), we get the new set of $n$ unit vectors $\{\mathbf{x}_1^{(n-1)},\cdots,\ \mathbf{x}_{n}^{(n-1)}\}$ with the same angle $\theta_{n-1}$ enclosed between any two of them where $\cos\theta_{n-1}=\frac{\cos\theta_n-\cos^2\theta_n}{1-\cos^2\theta_n}\\\implies \cos\theta_{n-1}=\frac{\cos\theta_n}{1+\cos\theta_n}$ since $\cos\theta_n\ne 0$ . Rearranging, we get the recursion, $\sec\theta_n=\sec\theta_{n-1}-1,\ n\ge 1$ Finally, noting that for $n=1$ , $\theta_1=\pi$ , as per definition, we get, $\cos\theta_n=-1/n$ .

Hence, the desired answer is $\cos\theta_4=\boxed{-0.25}$ and letting $n\to \infty, \ \theta_{n}\to\boxed{\pi/2}$

Looking at the other solutions here, I'm feeling a little (a lot) overwhelmed. I solved this in a slightly different way(might be equivalent to @Abhishek Sinha 's solution - I'm not sure if the solution is correct, so please tell me if anything's wrong. Let's go.

Let's make the $n + 1$ vectors have the same magnitude of 1, with no loss of generality. This doesn't change the angle between any two vectors, which is still $\cos(\theta_n)$ .
I claim that the sum of these vectors must be a zero vector, because of the symmetry of the construction means the vector cannot "point" in a specific direction. (If someone can provide a rigorous proof, I'll be happy to include it.)

Okay, so let's take the entire construction, and rotate it so that one of the vectors lies along one of the n dimensional axes. Let's name this vector $\hat{i_1}$ because it's neat.
Now, let's name the rest of the vectors too, keeping $\hat{i_1}, \hat{i_1}, ..., \hat{i_n}$ as the unit vectors of our space, as: ( $k$ is not the letter, it's the component of the vector along the $n^{th}$ dimension.)

$\\ \vec{v_1} = a_1 \hat{i_1} + b_1 \hat{i_2} + ... + k_1 \hat{i_n}$ $\\ \vec{v_2} = a_2 \hat{i_1} + b_2 \hat{i_2} + ... + k_2 \hat{i_n}$ $\\ ...$ $\\ \vec{v_n} = a_n \hat{i_1} + b_n \hat{i_2} + ... + k_n \hat{i_n}$

Note that $i_j \cdot i_k = 0$ for $j \neq k$ , where $\cdot$ is the dot product. This is because they are perpendicular in the n-dimensional space. (Rigorous proof encouraged.)
But, by the definition of the dot product, $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos(\theta)$ , where $\theta$ is the angle between $\vec{b}$ and $\vec{a}$ .

Back to the question now, we have, taking dot products of the vectors with $\hat{i_1}$ , $\cos(\theta_n) = \dfrac{\hat{i_1} \cdot \vec{v_1}}{|\hat{i_1}||\vec{v_1}|} = ... = \dfrac{\hat{i_1} \cdot \vec{v_n}}{|\hat{i_1}||\vec{v_n}|}$

$|\hat{i_1}| = |\vec{v_1}| = ... = |\vec{v_n}| = 1 \Rightarrow \cos(\theta_n) = a_1 = a_2 = ... = a_n$

But $\hat{i_1} + a_1 \hat{i_1} + ... + a_n \hat{i_1} = (1 + a_1 + ... + a_n ) \hat{i_1} = 0$
from the zero vector condition above (along one dimension).
This means, $n \cos(\theta_n) + 1 = 0 \Rightarrow \cos(\theta_n) = \frac{-1}{n}$
Here, $n = 4 \Rightarrow \cos(\theta_n) = -0.25$ .
That's it.

The solution I had in mind is similar to that of Abhishek. I assume that the reader has taken a first course in Linear Algebra. If you have not yet done so, you should! ;)

Let $\mathbf{u}_1,...,\mathbf{u}_{n+1}$ be unit vectors in $\mathbb{R}^n$ with $\mathbf{u}_i\cdot\mathbf{u}_j=\cos(\theta_n)$ , for $i\neq j$ . Let $\mathbf{G}=(\mathbf{u}_i\cdot\mathbf{u}_j)_{ij}$ be the associated Gram matrix, which is singular and positive semidefinite, so that the smallest eigenvalue of $\mathbf{G}$ is 0. Note that the diagonal entries of $\mathbf{G}$ are 1, while all other entries are $\cos(\theta_n)$ . As we discussed here , the eigenvalues of $\mathbf{G}$ will be $1-\cos(\theta_n)>0$ and $n\cos(\theta_n)+1=0$ , so that $\cos(\theta_n)=-\frac{1}{n}$ .

In particular, $\cos(\theta_4)=\boxed{-0.25}$