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I suppose that the standard $3$-simplex must be defined in $\mathbb{R}^4$ because you need four independientes unit vectors.

But, since a $n$-simplex is generate by $n+1$ affinely independient points $p_0,\dots,p_n$, I think that in $\mathbb{R}^3$ you can describe three coplanar points and another one "external" to them, so you get a thetaedron ($3$-simplex) in $\mathbb{R}^3$.

But the standard one is in $\mathbb{R}^4$. How is this possible?

Thanks

user183002
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    It's not very clear what you are asking. Any four points in $\Bbb{R}^3$ that are not coplanar span a $3$-simplex. However a $3$-simplex in $\Bbb{R}^3$ does not enjoy many of the nice properties of the standard simplex in $\Bbb{R}^4$, e.g. in $\Bbb{R}^3$ a $3$-simplex cannot have all its edges of the same length. – Rob Arthan Mar 03 '20 at 21:06
  • A $3$-simplex is just a tetrahedron, which is a $3$-dimensional object. Generally, an $n$-simplex is an $n$-dimensional object, and thus requires the spatial dimensions to be at least $n$. – md2perpe Mar 03 '20 at 21:09
  • Rob Arthan, why can't have all edges of same length? You ever can take points at equal distance among them, no? – user183002 Mar 03 '20 at 21:30
  • For example a regular thetaedron – user183002 Mar 03 '20 at 21:36
  • There is an affine map between the convex hull of $0,e_1,e_2,e_3$ in $\mathbb{R}^3$ and the standard simplex in $\mathbb{R}^4$. In barycentric coordinates one such map is the indentity. – copper.hat Mar 03 '20 at 21:42
  • copper.hat, can you answer my doubt? Why can't you get a thetaedron in $\mathbb{R}^3$ with edges of same length? – user183002 Mar 03 '20 at 21:47

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