Can a set of tetrahedra glued together by a common vertex be isometrically embedded in R^4?

A collection of triangles with a common vertex $A_1VA_2$, $A_2VA_3$, ... $A_NVA_1$ with specified side lengths can be isometrically embedded in $R^2$ provided the angles around $V$ add up to $2\pi$. If they do not however, I believe this can still be embedded in $R^3$ with one exception: $N=3$ and sum of the angles $>2\pi$. My question is: does this generalize to tetrahedra embedded in $R^4$? I expect that it does with a possible exception of 4 tetrahedra with solid angles adding to $>4\pi$. Would the answer be any different if the euclidian $R^4$ was replaced with $R^4$ with Minkowski metric (-,+,+,+)? Can you point me to some literature on the subject?

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Your belief is false: if any one of the angles is greater than the sum of all the others, then the triangles can't be embedded in $\mathbb{R}^3$.

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The embeddings of a collection of triangles with a common vertex is the same as an embedding of a graph in $S^2$.
The same way, the embedding of a collection of tetrahedra with a common vertex can be reformulated as embedding spherical polyhedral space into $S^3$. The embeddings into $\mathbb R^{3,1}$ correspond to embeddings of hyperbolic polyhedral space into Lobachevsky 3-space.
I think the most interesting theorem about this is Alexandrov's embedding theorem --- the theorem usually formulated for embedding of polyhedral spaces in to $\mathbb R^3$, but it was also proved for embedding of spherical polyhedral spaces to $S^3$ as well as for embedding hyperbolic polyhedral space into Lobachevsky 3-space.