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There are geometric and topological properties that significantly distinguish between Cartesian spaces of odd and even dimensions. Take, for example, the "sphere combing" property. The $(n-1)$-dimensional sphere (the boundary of a ball) in $\mathbb{R}^n$ admits a non-vanishing continuous tangent vector field if and only if $n$ is even.

Another striking difference is related to the existence of Hadamard matrices. It is known that an $n\times n$ Hadamard matrix exists only if $n=1,\ 2$, or is a multiple of $4$ (see https://en.wikipedia.org/wiki/Hadamard_matrix ), and Hadamard conjectured that there exists such a matrix for every $n$ divisible by $4$. The conjecture is still unsolved, but there exist many constructions producing an $n\times n$ Hadamard matrix for infinitely many values of $n$. Now, the geometric connection is this: Some $n+1$ vertices of the $n$-dimensional cube form a regular simplex if anand only if there exists an $(n+1)\times(n+1)$ Hadamard matrix.

Thus, spaces $\mathbb{R}^n$ and $\mathbb{R}^{n+1}$ often differ essentially for infinitely many values of $n$, and I think many other examples of this nature can be readily presented.

There are geometric and topological properties that significantly distinguish between Cartesian spaces of odd and even dimensions. Take, for example, the "sphere combing" property. The $(n-1)$-dimensional sphere (the boundary of a ball) in $\mathbb{R}^n$ admits a non-vanishing continuous tangent vector field if and only if $n$ is even.

Another striking difference is related to the existence of Hadamard matrices. It is known that an $n\times n$ Hadamard matrix exists only if $n=1,\ 2$, or is a multiple of $4$ (see https://en.wikipedia.org/wiki/Hadamard_matrix ), and Hadamard conjectured that there exists such a matrix for every $n$ divisible by $4$. The conjecture is still unsolved, but there exist many constructions producing an $n\times n$ Hadamard matrix for infinitely many values of $n$. Now, the geometric connection is this: Some $n+1$ vertices of the $n$-dimensional cube form a regular simplex if an only if there exists an $(n+1)\times(n+1)$ Hadamard matrix.

Thus, spaces $\mathbb{R}^n$ and $\mathbb{R}^{n+1}$ often differ essentially for infinitely many values of $n$, and I think many other examples of this nature can be readily presented.

There are geometric and topological properties that significantly distinguish between Cartesian spaces of odd and even dimensions. Take, for example, the "sphere combing" property. The $(n-1)$-dimensional sphere (the boundary of a ball) in $\mathbb{R}^n$ admits a non-vanishing continuous tangent vector field if and only if $n$ is even.

Another striking difference is related to the existence of Hadamard matrices. It is known that an $n\times n$ Hadamard matrix exists only if $n=1,\ 2$, or is a multiple of $4$ (see https://en.wikipedia.org/wiki/Hadamard_matrix ), and Hadamard conjectured that there exists such a matrix for every $n$ divisible by $4$. The conjecture is still unsolved, but there exist many constructions producing an $n\times n$ Hadamard matrix for infinitely many values of $n$. Now, the geometric connection is this: Some $n+1$ vertices of the $n$-dimensional cube form a regular simplex if and only if there exists an $(n+1)\times(n+1)$ Hadamard matrix.

Thus, spaces $\mathbb{R}^n$ and $\mathbb{R}^{n+1}$ often differ essentially for infinitely many values of $n$, and I think many other examples of this nature can be readily presented.

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There are geometric and topological properties that significantly distinguish between Cartesian spaces of odd and even dimensions. Take, for example, the "sphere combing" property. The $(n-1)$-dimensional sphere (the boundary of a ball) in $\mathbb{R}^n$ admits a non-vanishing continuous tangent vector field if and only if $n$ is even.

Another striking difference is related to the existence of Hadamard matrices. It is known that an $n\times n$ Hadamard matrix exists only if $n=1,\ 2$, or is a multiple of $4$ (see https://en.wikipedia.org/wiki/Hadamard_matrix ), and Hadamard conjectured that there exists such a matrix for every $n$ divisible by $4$. The conjecture is still unsolved, but there exist many constructions producing an $n\times n$ Hadamard matrix for infinitely many values of $n$. Now, the geometric connection is this: Some $n+1$ vertices of the $n$-dimensional cube form a regular simplex if an only if there exists an $(n+1)\times(n+1)$ Hadamard matrix.

Thus, spaces $\mathbb{R}^n$ and $\mathbb{R}^{n+1}$ often differ essentially for infinitely many values of $n$, and I think many other examples of this nature can be readily presented.

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