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Are there mathematical concepts that exist in the fourth dimension, but not in the third dimension?


How Do You Know If Mathematical Definition Matches Up With Reality?When is a function a dimension?Is there a specific mathematical term for a shape whose dimensions are defined?What does it mean to solve an equation?Does there exist a finite false proposition $P$ such that assuming $P$ we can derive any given false proposition?What does it mean when we say a mathematical object exists?






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Are there mathematical concepts that exist in the fourth dimension, but not in the third dimension? Of course, mathematical concepts include geometrical concepts, but I don't mean to say geometrical concept exclusively. I am not a mathematician and I am more of a layman, so it would be appreciated if you could tell what the concepts are in your answer so that a layman can understand.










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  • 1




    $begingroup$
    Not sure how to explain it in layman's terms, but the regular octaplex is totally unique to 4 dimensions.
    $endgroup$
    – eyeballfrog
    10 hours ago






  • 5




    $begingroup$
    What about quaternions?
    $endgroup$
    – J. W. Tanner
    9 hours ago

















6












$begingroup$


Are there mathematical concepts that exist in the fourth dimension, but not in the third dimension? Of course, mathematical concepts include geometrical concepts, but I don't mean to say geometrical concept exclusively. I am not a mathematician and I am more of a layman, so it would be appreciated if you could tell what the concepts are in your answer so that a layman can understand.










share|cite|improve this question







New contributor



jojafett is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$









  • 1




    $begingroup$
    Not sure how to explain it in layman's terms, but the regular octaplex is totally unique to 4 dimensions.
    $endgroup$
    – eyeballfrog
    10 hours ago






  • 5




    $begingroup$
    What about quaternions?
    $endgroup$
    – J. W. Tanner
    9 hours ago













6












6








6





$begingroup$


Are there mathematical concepts that exist in the fourth dimension, but not in the third dimension? Of course, mathematical concepts include geometrical concepts, but I don't mean to say geometrical concept exclusively. I am not a mathematician and I am more of a layman, so it would be appreciated if you could tell what the concepts are in your answer so that a layman can understand.










share|cite|improve this question







New contributor



jojafett is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




Are there mathematical concepts that exist in the fourth dimension, but not in the third dimension? Of course, mathematical concepts include geometrical concepts, but I don't mean to say geometrical concept exclusively. I am not a mathematician and I am more of a layman, so it would be appreciated if you could tell what the concepts are in your answer so that a layman can understand.







geometry philosophy






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  • 1




    $begingroup$
    Not sure how to explain it in layman's terms, but the regular octaplex is totally unique to 4 dimensions.
    $endgroup$
    – eyeballfrog
    10 hours ago






  • 5




    $begingroup$
    What about quaternions?
    $endgroup$
    – J. W. Tanner
    9 hours ago












  • 1




    $begingroup$
    Not sure how to explain it in layman's terms, but the regular octaplex is totally unique to 4 dimensions.
    $endgroup$
    – eyeballfrog
    10 hours ago






  • 5




    $begingroup$
    What about quaternions?
    $endgroup$
    – J. W. Tanner
    9 hours ago







1




1




$begingroup$
Not sure how to explain it in layman's terms, but the regular octaplex is totally unique to 4 dimensions.
$endgroup$
– eyeballfrog
10 hours ago




$begingroup$
Not sure how to explain it in layman's terms, but the regular octaplex is totally unique to 4 dimensions.
$endgroup$
– eyeballfrog
10 hours ago




5




5




$begingroup$
What about quaternions?
$endgroup$
– J. W. Tanner
9 hours ago




$begingroup$
What about quaternions?
$endgroup$
– J. W. Tanner
9 hours ago










3 Answers
3






active

oldest

votes


















12














$begingroup$

The one that sticks out for me the most is that there are five regular polytopes (called Platonic solids) in $3$ dimensions, and they all have analogues in $4$ dimensions, but there is another regular polytope in $4$ dimensions: the 24 cell.



The kicker is that in dimensions higher than $5$... there are only three regular polytopes!




Another thing that can happen in $4$ dimensional space but not $3$ is that you can have two planes which only intersect at the origin (and nowhere else.) In $3$ dimensions you'd get at least a line in the intersection.




I don't know if this also counts, but linear transformations in $3$-dimensions always stretch one direction (that is, they have a real eigenvector). This means that in all cases, a line in one direction must either stay put or be reversed to lie upon itself. In $4$ dimensions, it's possible to have transformations (even nonsingular ones) that don't have any real eigenvectors, so all lines get shifted.




Also not sure if this counts, but there are no $3$ dimensional asociative algebras over $mathbb R$ which allow division (they're called division algebras) but there is a unique $4$ dimensional one. (Look up the Frobenius theorem






share|cite|improve this answer











$endgroup$






















    4














    $begingroup$

    The fact that $mathbbR^4$ can be given a multiplication such that with vector's addition, it has a structure of (non commutative) field, i.e. quaternions.



    Whereas one can prove that it is impossible for $mathbbR^3$ to have a vector multiplication such that with usual addition of vectors, $mathbbR^3$ is a field (think for example to an unsuccessfull candidate, cross product, which is not even associative).



    I just saw now that at the end of his answer @rschwieb has mentionned "division algebras" ; but the concept of "field" is more powerful.






    share|cite|improve this answer











    $endgroup$














    • $begingroup$
      A non-commutative field is the same as a division ring. A division ring with a compatible vector space structure is the same as a division algera.
      $endgroup$
      – tomasz
      58 mins ago


















    1














    $begingroup$

    The fourth dimension is, in some ways, very peculiar.



    The $3$-dimensional Euclidean space has a unique differentiable structure. In fact, the same is true in any dimension not equal to $4$. By contrast, the $4$-dimensional space has a continuum of incompatible differentiable structures. Intuitively, this means that something that looks like the familiar $4$-dimensional space can have a large variety of rather odd geometries.



    An at most $3$-dimensional sphere admits a unique differentiable structure (there are no exotic spheres).



    It is not known whether there are exotic $4$-dimensional spheres, or whether there are finitely many of them (in contrast, in any other dimension, there are only finitely many differentiable structures on the sphere --- e.g. $28$ in $7$ dimensions).






    share|cite|improve this answer











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      3 Answers
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      3 Answers
      3






      active

      oldest

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      active

      oldest

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      active

      oldest

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      12














      $begingroup$

      The one that sticks out for me the most is that there are five regular polytopes (called Platonic solids) in $3$ dimensions, and they all have analogues in $4$ dimensions, but there is another regular polytope in $4$ dimensions: the 24 cell.



      The kicker is that in dimensions higher than $5$... there are only three regular polytopes!




      Another thing that can happen in $4$ dimensional space but not $3$ is that you can have two planes which only intersect at the origin (and nowhere else.) In $3$ dimensions you'd get at least a line in the intersection.




      I don't know if this also counts, but linear transformations in $3$-dimensions always stretch one direction (that is, they have a real eigenvector). This means that in all cases, a line in one direction must either stay put or be reversed to lie upon itself. In $4$ dimensions, it's possible to have transformations (even nonsingular ones) that don't have any real eigenvectors, so all lines get shifted.




      Also not sure if this counts, but there are no $3$ dimensional asociative algebras over $mathbb R$ which allow division (they're called division algebras) but there is a unique $4$ dimensional one. (Look up the Frobenius theorem






      share|cite|improve this answer











      $endgroup$



















        12














        $begingroup$

        The one that sticks out for me the most is that there are five regular polytopes (called Platonic solids) in $3$ dimensions, and they all have analogues in $4$ dimensions, but there is another regular polytope in $4$ dimensions: the 24 cell.



        The kicker is that in dimensions higher than $5$... there are only three regular polytopes!




        Another thing that can happen in $4$ dimensional space but not $3$ is that you can have two planes which only intersect at the origin (and nowhere else.) In $3$ dimensions you'd get at least a line in the intersection.




        I don't know if this also counts, but linear transformations in $3$-dimensions always stretch one direction (that is, they have a real eigenvector). This means that in all cases, a line in one direction must either stay put or be reversed to lie upon itself. In $4$ dimensions, it's possible to have transformations (even nonsingular ones) that don't have any real eigenvectors, so all lines get shifted.




        Also not sure if this counts, but there are no $3$ dimensional asociative algebras over $mathbb R$ which allow division (they're called division algebras) but there is a unique $4$ dimensional one. (Look up the Frobenius theorem






        share|cite|improve this answer











        $endgroup$

















          12














          12










          12







          $begingroup$

          The one that sticks out for me the most is that there are five regular polytopes (called Platonic solids) in $3$ dimensions, and they all have analogues in $4$ dimensions, but there is another regular polytope in $4$ dimensions: the 24 cell.



          The kicker is that in dimensions higher than $5$... there are only three regular polytopes!




          Another thing that can happen in $4$ dimensional space but not $3$ is that you can have two planes which only intersect at the origin (and nowhere else.) In $3$ dimensions you'd get at least a line in the intersection.




          I don't know if this also counts, but linear transformations in $3$-dimensions always stretch one direction (that is, they have a real eigenvector). This means that in all cases, a line in one direction must either stay put or be reversed to lie upon itself. In $4$ dimensions, it's possible to have transformations (even nonsingular ones) that don't have any real eigenvectors, so all lines get shifted.




          Also not sure if this counts, but there are no $3$ dimensional asociative algebras over $mathbb R$ which allow division (they're called division algebras) but there is a unique $4$ dimensional one. (Look up the Frobenius theorem






          share|cite|improve this answer











          $endgroup$



          The one that sticks out for me the most is that there are five regular polytopes (called Platonic solids) in $3$ dimensions, and they all have analogues in $4$ dimensions, but there is another regular polytope in $4$ dimensions: the 24 cell.



          The kicker is that in dimensions higher than $5$... there are only three regular polytopes!




          Another thing that can happen in $4$ dimensional space but not $3$ is that you can have two planes which only intersect at the origin (and nowhere else.) In $3$ dimensions you'd get at least a line in the intersection.




          I don't know if this also counts, but linear transformations in $3$-dimensions always stretch one direction (that is, they have a real eigenvector). This means that in all cases, a line in one direction must either stay put or be reversed to lie upon itself. In $4$ dimensions, it's possible to have transformations (even nonsingular ones) that don't have any real eigenvectors, so all lines get shifted.




          Also not sure if this counts, but there are no $3$ dimensional asociative algebras over $mathbb R$ which allow division (they're called division algebras) but there is a unique $4$ dimensional one. (Look up the Frobenius theorem







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 9 hours ago

























          answered 10 hours ago









          rschwiebrschwieb

          113k12 gold badges112 silver badges266 bronze badges




          113k12 gold badges112 silver badges266 bronze badges


























              4














              $begingroup$

              The fact that $mathbbR^4$ can be given a multiplication such that with vector's addition, it has a structure of (non commutative) field, i.e. quaternions.



              Whereas one can prove that it is impossible for $mathbbR^3$ to have a vector multiplication such that with usual addition of vectors, $mathbbR^3$ is a field (think for example to an unsuccessfull candidate, cross product, which is not even associative).



              I just saw now that at the end of his answer @rschwieb has mentionned "division algebras" ; but the concept of "field" is more powerful.






              share|cite|improve this answer











              $endgroup$














              • $begingroup$
                A non-commutative field is the same as a division ring. A division ring with a compatible vector space structure is the same as a division algera.
                $endgroup$
                – tomasz
                58 mins ago















              4














              $begingroup$

              The fact that $mathbbR^4$ can be given a multiplication such that with vector's addition, it has a structure of (non commutative) field, i.e. quaternions.



              Whereas one can prove that it is impossible for $mathbbR^3$ to have a vector multiplication such that with usual addition of vectors, $mathbbR^3$ is a field (think for example to an unsuccessfull candidate, cross product, which is not even associative).



              I just saw now that at the end of his answer @rschwieb has mentionned "division algebras" ; but the concept of "field" is more powerful.






              share|cite|improve this answer











              $endgroup$














              • $begingroup$
                A non-commutative field is the same as a division ring. A division ring with a compatible vector space structure is the same as a division algera.
                $endgroup$
                – tomasz
                58 mins ago













              4














              4










              4







              $begingroup$

              The fact that $mathbbR^4$ can be given a multiplication such that with vector's addition, it has a structure of (non commutative) field, i.e. quaternions.



              Whereas one can prove that it is impossible for $mathbbR^3$ to have a vector multiplication such that with usual addition of vectors, $mathbbR^3$ is a field (think for example to an unsuccessfull candidate, cross product, which is not even associative).



              I just saw now that at the end of his answer @rschwieb has mentionned "division algebras" ; but the concept of "field" is more powerful.






              share|cite|improve this answer











              $endgroup$



              The fact that $mathbbR^4$ can be given a multiplication such that with vector's addition, it has a structure of (non commutative) field, i.e. quaternions.



              Whereas one can prove that it is impossible for $mathbbR^3$ to have a vector multiplication such that with usual addition of vectors, $mathbbR^3$ is a field (think for example to an unsuccessfull candidate, cross product, which is not even associative).



              I just saw now that at the end of his answer @rschwieb has mentionned "division algebras" ; but the concept of "field" is more powerful.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 1 hour ago

























              answered 2 hours ago









              Jean MarieJean Marie

              35.4k4 gold badges26 silver badges60 bronze badges




              35.4k4 gold badges26 silver badges60 bronze badges














              • $begingroup$
                A non-commutative field is the same as a division ring. A division ring with a compatible vector space structure is the same as a division algera.
                $endgroup$
                – tomasz
                58 mins ago
















              • $begingroup$
                A non-commutative field is the same as a division ring. A division ring with a compatible vector space structure is the same as a division algera.
                $endgroup$
                – tomasz
                58 mins ago















              $begingroup$
              A non-commutative field is the same as a division ring. A division ring with a compatible vector space structure is the same as a division algera.
              $endgroup$
              – tomasz
              58 mins ago




              $begingroup$
              A non-commutative field is the same as a division ring. A division ring with a compatible vector space structure is the same as a division algera.
              $endgroup$
              – tomasz
              58 mins ago











              1














              $begingroup$

              The fourth dimension is, in some ways, very peculiar.



              The $3$-dimensional Euclidean space has a unique differentiable structure. In fact, the same is true in any dimension not equal to $4$. By contrast, the $4$-dimensional space has a continuum of incompatible differentiable structures. Intuitively, this means that something that looks like the familiar $4$-dimensional space can have a large variety of rather odd geometries.



              An at most $3$-dimensional sphere admits a unique differentiable structure (there are no exotic spheres).



              It is not known whether there are exotic $4$-dimensional spheres, or whether there are finitely many of them (in contrast, in any other dimension, there are only finitely many differentiable structures on the sphere --- e.g. $28$ in $7$ dimensions).






              share|cite|improve this answer











              $endgroup$



















                1














                $begingroup$

                The fourth dimension is, in some ways, very peculiar.



                The $3$-dimensional Euclidean space has a unique differentiable structure. In fact, the same is true in any dimension not equal to $4$. By contrast, the $4$-dimensional space has a continuum of incompatible differentiable structures. Intuitively, this means that something that looks like the familiar $4$-dimensional space can have a large variety of rather odd geometries.



                An at most $3$-dimensional sphere admits a unique differentiable structure (there are no exotic spheres).



                It is not known whether there are exotic $4$-dimensional spheres, or whether there are finitely many of them (in contrast, in any other dimension, there are only finitely many differentiable structures on the sphere --- e.g. $28$ in $7$ dimensions).






                share|cite|improve this answer











                $endgroup$

















                  1














                  1










                  1







                  $begingroup$

                  The fourth dimension is, in some ways, very peculiar.



                  The $3$-dimensional Euclidean space has a unique differentiable structure. In fact, the same is true in any dimension not equal to $4$. By contrast, the $4$-dimensional space has a continuum of incompatible differentiable structures. Intuitively, this means that something that looks like the familiar $4$-dimensional space can have a large variety of rather odd geometries.



                  An at most $3$-dimensional sphere admits a unique differentiable structure (there are no exotic spheres).



                  It is not known whether there are exotic $4$-dimensional spheres, or whether there are finitely many of them (in contrast, in any other dimension, there are only finitely many differentiable structures on the sphere --- e.g. $28$ in $7$ dimensions).






                  share|cite|improve this answer











                  $endgroup$



                  The fourth dimension is, in some ways, very peculiar.



                  The $3$-dimensional Euclidean space has a unique differentiable structure. In fact, the same is true in any dimension not equal to $4$. By contrast, the $4$-dimensional space has a continuum of incompatible differentiable structures. Intuitively, this means that something that looks like the familiar $4$-dimensional space can have a large variety of rather odd geometries.



                  An at most $3$-dimensional sphere admits a unique differentiable structure (there are no exotic spheres).



                  It is not known whether there are exotic $4$-dimensional spheres, or whether there are finitely many of them (in contrast, in any other dimension, there are only finitely many differentiable structures on the sphere --- e.g. $28$ in $7$ dimensions).







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 56 mins ago

























                  answered 1 hour ago









                  tomasztomasz

                  24.6k2 gold badges36 silver badges83 bronze badges




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