How do we know you can only flip something’s orientation two times?











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I was exploring what the determinant’s sign means geometrically. For 2-D, you can swap the axes and you’ve flipped orientation, and once you swap them again, you get the original orientation.



Why should this be the case for higher dimensions? Why is it not possible that in 3-D, I can swap i and j, then swap j and k, and I get a “third orientation”?



How is orientation defined anyway?












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    up vote
    8
    down vote

    favorite












    I was exploring what the determinant’s sign means geometrically. For 2-D, you can swap the axes and you’ve flipped orientation, and once you swap them again, you get the original orientation.



    Why should this be the case for higher dimensions? Why is it not possible that in 3-D, I can swap i and j, then swap j and k, and I get a “third orientation”?



    How is orientation defined anyway?












    share|cite|improve this question


























      up vote
      8
      down vote

      favorite









      up vote
      8
      down vote

      favorite











      I was exploring what the determinant’s sign means geometrically. For 2-D, you can swap the axes and you’ve flipped orientation, and once you swap them again, you get the original orientation.



      Why should this be the case for higher dimensions? Why is it not possible that in 3-D, I can swap i and j, then swap j and k, and I get a “third orientation”?



      How is orientation defined anyway?












      share|cite|improve this question















      I was exploring what the determinant’s sign means geometrically. For 2-D, you can swap the axes and you’ve flipped orientation, and once you swap them again, you get the original orientation.



      Why should this be the case for higher dimensions? Why is it not possible that in 3-D, I can swap i and j, then swap j and k, and I get a “third orientation”?



      How is orientation defined anyway?









      linear-algebra






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 19 at 2:26

























      asked Nov 19 at 1:25









      Yatharth Agarwal

      471417




      471417






















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          3
          down vote













          Here is a geometric reason for the definition of orientation.



          The set of all invertible $n times n$ matrices forms a group denoted $GL(n,mathbb R)$, and it is identified with the set of all invertible linear transformations $T : mathbb R^n to mathbb R^n$.



          This group decomposes into two subsets, which I'll denote
          $$GL(n,mathbb R) = {GL}_+(n,mathbb R) cup {GL}_-(n,mathbb R)
          $$

          where the subscript represents the sign of the determinant.



          Here are some key algebraic facts (easily derived from the theorem that the determinant of a product equals the product of the determinants):




          • The product of any two matrices of positive determinant is another matrix of positive determinant. In other words, ${GL}_+(n,mathbb R)$ is a subgroup of $GL(n,mathbb R)$.

          • The product of any two matrices of negative determinant is a matrix of positive determinant.

          • The product of a matrix of positive determinant with a matrix of negative determinant (in either order) is a matrix of negative determinant.


          These three algebraic facts can be summarized in a single statement: there is a group homomorphism $sigma : GL(n,mathbb R) to {-1,+1}$ which assigns $+1$ to a matrix of positive determinant and $-1$ to a matrix of negative determinant.



          Here are two key geometric facts:





          • Theorem: Any two matrices in ${GL}_+(n,mathbb R)$ can be connected by a continuous path of matrices in ${GL}+(n,mathbb R)$.


          • Corollary: Also, any two matrices in ${GL}_-(n,mathbb R)$ can be connected by a continuous path of matrices in ${GL}_-(n,mathbb R)$.


          You asked:




          Why is it not possible that in 3-D, I can swap i and j, then swap j and k, and I get a “third orientation”?




          Because if you swap $i$ and $j$ using the following reflection matrix (which has negative determinant)
          $$R_1 = pmatrix{0&1&0\1&0&0\0&0&1}$$
          and after that if you then swap $j$ and $k$ using the following reflection matrix (which also has negative determinant)
          $$R_2 = pmatrix{1&0&0\0&0&1\0&1&0}
          $$

          and then you formulate the product matrix
          $$R_2 R_1 = pmatrix{0&1&0\0&0&1\1&0&0}
          $$

          then $R_2 R_1$ has positive determinant, as you can check algebraically. Therefore, once you have convinced yourself of the algebraic and geometric facts listed earlier, you will be confident that $R_2 R_1$ does not represent a "new orientation" because it is connected by a continuous path of matrices with every other positive determinant matrix, i.e. with every rotation matrix and with the identity itself.



          In fact, and you can verify this by calculation, $R_2 R_1$ is itself a rotation matrix: it is a rotation by $2pi/3$ around the axis which is parameterized by $t mapsto (t,t,t)$. If you fix this axis, and then vary the angle of rotation continuously, starting from the angle $2pi/3$, and varying it down to the angle $0$, you will have produced a continuous path which connects the matrix $R_2 R_1$ with the identity matrix.



          To summarize, the reason there is not a "third orientation" is because orientation is defined so as to summarize, and to be consistent with, the key algebraic and geometric facts that are listed above. In other words: invertible $n times n$ matrices are naturally subdivided into two classes, and we choose to describe those two classes using the terminology of "orientations".



          One last word: If you've not seen that theorem before, digging into its proof would be an excellent way to understand the geometric reasons why orientation is defined the way that it is.






          share|cite|improve this answer






























            up vote
            1
            down vote













            A reasonable question, indeed, on geometric grounds.



            If you track the details of your swap-$i$-and-$j$, and then $j$-and-$k$, you'll see that that composite action is a rotation (preserving orientation).



            More generally, the composition of two reflections (the simplest orientation-reversing isometries) is a rotation (orientation-preserving). You can work out the details either geometrically or algebraically.






            share|cite|improve this answer





















            • More generally, the composition of two orientation reversing isometries is orientation preserving.
              – Matt Samuel
              Nov 19 at 1:35










            • Could I ask: How is orientation defined, as you use it?
              – Yatharth Agarwal
              Nov 19 at 2:24










            • Orientation is the sign of the determinant of the linear transformation (i.e. the basis change matrix): $+$ sign is orientation preserving; $-$ sign is orientation reversing.
              – Lee Mosher
              Nov 19 at 2:34










            • @LeeMosher That’s a little circular for me, given I’m trying to understand what the sign of the determinant means geometrically. Are there alternative interpretations/definitions?
              – Yatharth Agarwal
              yesterday










            • Your quest to understand orientation geometrically is a good one, but, my comment was simply an answer to your question in the preceding comment.
              – Lee Mosher
              12 hours ago











            Your Answer





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            2 Answers
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            2 Answers
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            up vote
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            down vote













            Here is a geometric reason for the definition of orientation.



            The set of all invertible $n times n$ matrices forms a group denoted $GL(n,mathbb R)$, and it is identified with the set of all invertible linear transformations $T : mathbb R^n to mathbb R^n$.



            This group decomposes into two subsets, which I'll denote
            $$GL(n,mathbb R) = {GL}_+(n,mathbb R) cup {GL}_-(n,mathbb R)
            $$

            where the subscript represents the sign of the determinant.



            Here are some key algebraic facts (easily derived from the theorem that the determinant of a product equals the product of the determinants):




            • The product of any two matrices of positive determinant is another matrix of positive determinant. In other words, ${GL}_+(n,mathbb R)$ is a subgroup of $GL(n,mathbb R)$.

            • The product of any two matrices of negative determinant is a matrix of positive determinant.

            • The product of a matrix of positive determinant with a matrix of negative determinant (in either order) is a matrix of negative determinant.


            These three algebraic facts can be summarized in a single statement: there is a group homomorphism $sigma : GL(n,mathbb R) to {-1,+1}$ which assigns $+1$ to a matrix of positive determinant and $-1$ to a matrix of negative determinant.



            Here are two key geometric facts:





            • Theorem: Any two matrices in ${GL}_+(n,mathbb R)$ can be connected by a continuous path of matrices in ${GL}+(n,mathbb R)$.


            • Corollary: Also, any two matrices in ${GL}_-(n,mathbb R)$ can be connected by a continuous path of matrices in ${GL}_-(n,mathbb R)$.


            You asked:




            Why is it not possible that in 3-D, I can swap i and j, then swap j and k, and I get a “third orientation”?




            Because if you swap $i$ and $j$ using the following reflection matrix (which has negative determinant)
            $$R_1 = pmatrix{0&1&0\1&0&0\0&0&1}$$
            and after that if you then swap $j$ and $k$ using the following reflection matrix (which also has negative determinant)
            $$R_2 = pmatrix{1&0&0\0&0&1\0&1&0}
            $$

            and then you formulate the product matrix
            $$R_2 R_1 = pmatrix{0&1&0\0&0&1\1&0&0}
            $$

            then $R_2 R_1$ has positive determinant, as you can check algebraically. Therefore, once you have convinced yourself of the algebraic and geometric facts listed earlier, you will be confident that $R_2 R_1$ does not represent a "new orientation" because it is connected by a continuous path of matrices with every other positive determinant matrix, i.e. with every rotation matrix and with the identity itself.



            In fact, and you can verify this by calculation, $R_2 R_1$ is itself a rotation matrix: it is a rotation by $2pi/3$ around the axis which is parameterized by $t mapsto (t,t,t)$. If you fix this axis, and then vary the angle of rotation continuously, starting from the angle $2pi/3$, and varying it down to the angle $0$, you will have produced a continuous path which connects the matrix $R_2 R_1$ with the identity matrix.



            To summarize, the reason there is not a "third orientation" is because orientation is defined so as to summarize, and to be consistent with, the key algebraic and geometric facts that are listed above. In other words: invertible $n times n$ matrices are naturally subdivided into two classes, and we choose to describe those two classes using the terminology of "orientations".



            One last word: If you've not seen that theorem before, digging into its proof would be an excellent way to understand the geometric reasons why orientation is defined the way that it is.






            share|cite|improve this answer



























              up vote
              3
              down vote













              Here is a geometric reason for the definition of orientation.



              The set of all invertible $n times n$ matrices forms a group denoted $GL(n,mathbb R)$, and it is identified with the set of all invertible linear transformations $T : mathbb R^n to mathbb R^n$.



              This group decomposes into two subsets, which I'll denote
              $$GL(n,mathbb R) = {GL}_+(n,mathbb R) cup {GL}_-(n,mathbb R)
              $$

              where the subscript represents the sign of the determinant.



              Here are some key algebraic facts (easily derived from the theorem that the determinant of a product equals the product of the determinants):




              • The product of any two matrices of positive determinant is another matrix of positive determinant. In other words, ${GL}_+(n,mathbb R)$ is a subgroup of $GL(n,mathbb R)$.

              • The product of any two matrices of negative determinant is a matrix of positive determinant.

              • The product of a matrix of positive determinant with a matrix of negative determinant (in either order) is a matrix of negative determinant.


              These three algebraic facts can be summarized in a single statement: there is a group homomorphism $sigma : GL(n,mathbb R) to {-1,+1}$ which assigns $+1$ to a matrix of positive determinant and $-1$ to a matrix of negative determinant.



              Here are two key geometric facts:





              • Theorem: Any two matrices in ${GL}_+(n,mathbb R)$ can be connected by a continuous path of matrices in ${GL}+(n,mathbb R)$.


              • Corollary: Also, any two matrices in ${GL}_-(n,mathbb R)$ can be connected by a continuous path of matrices in ${GL}_-(n,mathbb R)$.


              You asked:




              Why is it not possible that in 3-D, I can swap i and j, then swap j and k, and I get a “third orientation”?




              Because if you swap $i$ and $j$ using the following reflection matrix (which has negative determinant)
              $$R_1 = pmatrix{0&1&0\1&0&0\0&0&1}$$
              and after that if you then swap $j$ and $k$ using the following reflection matrix (which also has negative determinant)
              $$R_2 = pmatrix{1&0&0\0&0&1\0&1&0}
              $$

              and then you formulate the product matrix
              $$R_2 R_1 = pmatrix{0&1&0\0&0&1\1&0&0}
              $$

              then $R_2 R_1$ has positive determinant, as you can check algebraically. Therefore, once you have convinced yourself of the algebraic and geometric facts listed earlier, you will be confident that $R_2 R_1$ does not represent a "new orientation" because it is connected by a continuous path of matrices with every other positive determinant matrix, i.e. with every rotation matrix and with the identity itself.



              In fact, and you can verify this by calculation, $R_2 R_1$ is itself a rotation matrix: it is a rotation by $2pi/3$ around the axis which is parameterized by $t mapsto (t,t,t)$. If you fix this axis, and then vary the angle of rotation continuously, starting from the angle $2pi/3$, and varying it down to the angle $0$, you will have produced a continuous path which connects the matrix $R_2 R_1$ with the identity matrix.



              To summarize, the reason there is not a "third orientation" is because orientation is defined so as to summarize, and to be consistent with, the key algebraic and geometric facts that are listed above. In other words: invertible $n times n$ matrices are naturally subdivided into two classes, and we choose to describe those two classes using the terminology of "orientations".



              One last word: If you've not seen that theorem before, digging into its proof would be an excellent way to understand the geometric reasons why orientation is defined the way that it is.






              share|cite|improve this answer

























                up vote
                3
                down vote










                up vote
                3
                down vote









                Here is a geometric reason for the definition of orientation.



                The set of all invertible $n times n$ matrices forms a group denoted $GL(n,mathbb R)$, and it is identified with the set of all invertible linear transformations $T : mathbb R^n to mathbb R^n$.



                This group decomposes into two subsets, which I'll denote
                $$GL(n,mathbb R) = {GL}_+(n,mathbb R) cup {GL}_-(n,mathbb R)
                $$

                where the subscript represents the sign of the determinant.



                Here are some key algebraic facts (easily derived from the theorem that the determinant of a product equals the product of the determinants):




                • The product of any two matrices of positive determinant is another matrix of positive determinant. In other words, ${GL}_+(n,mathbb R)$ is a subgroup of $GL(n,mathbb R)$.

                • The product of any two matrices of negative determinant is a matrix of positive determinant.

                • The product of a matrix of positive determinant with a matrix of negative determinant (in either order) is a matrix of negative determinant.


                These three algebraic facts can be summarized in a single statement: there is a group homomorphism $sigma : GL(n,mathbb R) to {-1,+1}$ which assigns $+1$ to a matrix of positive determinant and $-1$ to a matrix of negative determinant.



                Here are two key geometric facts:





                • Theorem: Any two matrices in ${GL}_+(n,mathbb R)$ can be connected by a continuous path of matrices in ${GL}+(n,mathbb R)$.


                • Corollary: Also, any two matrices in ${GL}_-(n,mathbb R)$ can be connected by a continuous path of matrices in ${GL}_-(n,mathbb R)$.


                You asked:




                Why is it not possible that in 3-D, I can swap i and j, then swap j and k, and I get a “third orientation”?




                Because if you swap $i$ and $j$ using the following reflection matrix (which has negative determinant)
                $$R_1 = pmatrix{0&1&0\1&0&0\0&0&1}$$
                and after that if you then swap $j$ and $k$ using the following reflection matrix (which also has negative determinant)
                $$R_2 = pmatrix{1&0&0\0&0&1\0&1&0}
                $$

                and then you formulate the product matrix
                $$R_2 R_1 = pmatrix{0&1&0\0&0&1\1&0&0}
                $$

                then $R_2 R_1$ has positive determinant, as you can check algebraically. Therefore, once you have convinced yourself of the algebraic and geometric facts listed earlier, you will be confident that $R_2 R_1$ does not represent a "new orientation" because it is connected by a continuous path of matrices with every other positive determinant matrix, i.e. with every rotation matrix and with the identity itself.



                In fact, and you can verify this by calculation, $R_2 R_1$ is itself a rotation matrix: it is a rotation by $2pi/3$ around the axis which is parameterized by $t mapsto (t,t,t)$. If you fix this axis, and then vary the angle of rotation continuously, starting from the angle $2pi/3$, and varying it down to the angle $0$, you will have produced a continuous path which connects the matrix $R_2 R_1$ with the identity matrix.



                To summarize, the reason there is not a "third orientation" is because orientation is defined so as to summarize, and to be consistent with, the key algebraic and geometric facts that are listed above. In other words: invertible $n times n$ matrices are naturally subdivided into two classes, and we choose to describe those two classes using the terminology of "orientations".



                One last word: If you've not seen that theorem before, digging into its proof would be an excellent way to understand the geometric reasons why orientation is defined the way that it is.






                share|cite|improve this answer














                Here is a geometric reason for the definition of orientation.



                The set of all invertible $n times n$ matrices forms a group denoted $GL(n,mathbb R)$, and it is identified with the set of all invertible linear transformations $T : mathbb R^n to mathbb R^n$.



                This group decomposes into two subsets, which I'll denote
                $$GL(n,mathbb R) = {GL}_+(n,mathbb R) cup {GL}_-(n,mathbb R)
                $$

                where the subscript represents the sign of the determinant.



                Here are some key algebraic facts (easily derived from the theorem that the determinant of a product equals the product of the determinants):




                • The product of any two matrices of positive determinant is another matrix of positive determinant. In other words, ${GL}_+(n,mathbb R)$ is a subgroup of $GL(n,mathbb R)$.

                • The product of any two matrices of negative determinant is a matrix of positive determinant.

                • The product of a matrix of positive determinant with a matrix of negative determinant (in either order) is a matrix of negative determinant.


                These three algebraic facts can be summarized in a single statement: there is a group homomorphism $sigma : GL(n,mathbb R) to {-1,+1}$ which assigns $+1$ to a matrix of positive determinant and $-1$ to a matrix of negative determinant.



                Here are two key geometric facts:





                • Theorem: Any two matrices in ${GL}_+(n,mathbb R)$ can be connected by a continuous path of matrices in ${GL}+(n,mathbb R)$.


                • Corollary: Also, any two matrices in ${GL}_-(n,mathbb R)$ can be connected by a continuous path of matrices in ${GL}_-(n,mathbb R)$.


                You asked:




                Why is it not possible that in 3-D, I can swap i and j, then swap j and k, and I get a “third orientation”?




                Because if you swap $i$ and $j$ using the following reflection matrix (which has negative determinant)
                $$R_1 = pmatrix{0&1&0\1&0&0\0&0&1}$$
                and after that if you then swap $j$ and $k$ using the following reflection matrix (which also has negative determinant)
                $$R_2 = pmatrix{1&0&0\0&0&1\0&1&0}
                $$

                and then you formulate the product matrix
                $$R_2 R_1 = pmatrix{0&1&0\0&0&1\1&0&0}
                $$

                then $R_2 R_1$ has positive determinant, as you can check algebraically. Therefore, once you have convinced yourself of the algebraic and geometric facts listed earlier, you will be confident that $R_2 R_1$ does not represent a "new orientation" because it is connected by a continuous path of matrices with every other positive determinant matrix, i.e. with every rotation matrix and with the identity itself.



                In fact, and you can verify this by calculation, $R_2 R_1$ is itself a rotation matrix: it is a rotation by $2pi/3$ around the axis which is parameterized by $t mapsto (t,t,t)$. If you fix this axis, and then vary the angle of rotation continuously, starting from the angle $2pi/3$, and varying it down to the angle $0$, you will have produced a continuous path which connects the matrix $R_2 R_1$ with the identity matrix.



                To summarize, the reason there is not a "third orientation" is because orientation is defined so as to summarize, and to be consistent with, the key algebraic and geometric facts that are listed above. In other words: invertible $n times n$ matrices are naturally subdivided into two classes, and we choose to describe those two classes using the terminology of "orientations".



                One last word: If you've not seen that theorem before, digging into its proof would be an excellent way to understand the geometric reasons why orientation is defined the way that it is.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 10 hours ago

























                answered 12 hours ago









                Lee Mosher

                47.5k33681




                47.5k33681






















                    up vote
                    1
                    down vote













                    A reasonable question, indeed, on geometric grounds.



                    If you track the details of your swap-$i$-and-$j$, and then $j$-and-$k$, you'll see that that composite action is a rotation (preserving orientation).



                    More generally, the composition of two reflections (the simplest orientation-reversing isometries) is a rotation (orientation-preserving). You can work out the details either geometrically or algebraically.






                    share|cite|improve this answer





















                    • More generally, the composition of two orientation reversing isometries is orientation preserving.
                      – Matt Samuel
                      Nov 19 at 1:35










                    • Could I ask: How is orientation defined, as you use it?
                      – Yatharth Agarwal
                      Nov 19 at 2:24










                    • Orientation is the sign of the determinant of the linear transformation (i.e. the basis change matrix): $+$ sign is orientation preserving; $-$ sign is orientation reversing.
                      – Lee Mosher
                      Nov 19 at 2:34










                    • @LeeMosher That’s a little circular for me, given I’m trying to understand what the sign of the determinant means geometrically. Are there alternative interpretations/definitions?
                      – Yatharth Agarwal
                      yesterday










                    • Your quest to understand orientation geometrically is a good one, but, my comment was simply an answer to your question in the preceding comment.
                      – Lee Mosher
                      12 hours ago















                    up vote
                    1
                    down vote













                    A reasonable question, indeed, on geometric grounds.



                    If you track the details of your swap-$i$-and-$j$, and then $j$-and-$k$, you'll see that that composite action is a rotation (preserving orientation).



                    More generally, the composition of two reflections (the simplest orientation-reversing isometries) is a rotation (orientation-preserving). You can work out the details either geometrically or algebraically.






                    share|cite|improve this answer





















                    • More generally, the composition of two orientation reversing isometries is orientation preserving.
                      – Matt Samuel
                      Nov 19 at 1:35










                    • Could I ask: How is orientation defined, as you use it?
                      – Yatharth Agarwal
                      Nov 19 at 2:24










                    • Orientation is the sign of the determinant of the linear transformation (i.e. the basis change matrix): $+$ sign is orientation preserving; $-$ sign is orientation reversing.
                      – Lee Mosher
                      Nov 19 at 2:34










                    • @LeeMosher That’s a little circular for me, given I’m trying to understand what the sign of the determinant means geometrically. Are there alternative interpretations/definitions?
                      – Yatharth Agarwal
                      yesterday










                    • Your quest to understand orientation geometrically is a good one, but, my comment was simply an answer to your question in the preceding comment.
                      – Lee Mosher
                      12 hours ago













                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    A reasonable question, indeed, on geometric grounds.



                    If you track the details of your swap-$i$-and-$j$, and then $j$-and-$k$, you'll see that that composite action is a rotation (preserving orientation).



                    More generally, the composition of two reflections (the simplest orientation-reversing isometries) is a rotation (orientation-preserving). You can work out the details either geometrically or algebraically.






                    share|cite|improve this answer












                    A reasonable question, indeed, on geometric grounds.



                    If you track the details of your swap-$i$-and-$j$, and then $j$-and-$k$, you'll see that that composite action is a rotation (preserving orientation).



                    More generally, the composition of two reflections (the simplest orientation-reversing isometries) is a rotation (orientation-preserving). You can work out the details either geometrically or algebraically.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 19 at 1:31









                    paul garrett

                    31.3k361117




                    31.3k361117












                    • More generally, the composition of two orientation reversing isometries is orientation preserving.
                      – Matt Samuel
                      Nov 19 at 1:35










                    • Could I ask: How is orientation defined, as you use it?
                      – Yatharth Agarwal
                      Nov 19 at 2:24










                    • Orientation is the sign of the determinant of the linear transformation (i.e. the basis change matrix): $+$ sign is orientation preserving; $-$ sign is orientation reversing.
                      – Lee Mosher
                      Nov 19 at 2:34










                    • @LeeMosher That’s a little circular for me, given I’m trying to understand what the sign of the determinant means geometrically. Are there alternative interpretations/definitions?
                      – Yatharth Agarwal
                      yesterday










                    • Your quest to understand orientation geometrically is a good one, but, my comment was simply an answer to your question in the preceding comment.
                      – Lee Mosher
                      12 hours ago


















                    • More generally, the composition of two orientation reversing isometries is orientation preserving.
                      – Matt Samuel
                      Nov 19 at 1:35










                    • Could I ask: How is orientation defined, as you use it?
                      – Yatharth Agarwal
                      Nov 19 at 2:24










                    • Orientation is the sign of the determinant of the linear transformation (i.e. the basis change matrix): $+$ sign is orientation preserving; $-$ sign is orientation reversing.
                      – Lee Mosher
                      Nov 19 at 2:34










                    • @LeeMosher That’s a little circular for me, given I’m trying to understand what the sign of the determinant means geometrically. Are there alternative interpretations/definitions?
                      – Yatharth Agarwal
                      yesterday










                    • Your quest to understand orientation geometrically is a good one, but, my comment was simply an answer to your question in the preceding comment.
                      – Lee Mosher
                      12 hours ago
















                    More generally, the composition of two orientation reversing isometries is orientation preserving.
                    – Matt Samuel
                    Nov 19 at 1:35




                    More generally, the composition of two orientation reversing isometries is orientation preserving.
                    – Matt Samuel
                    Nov 19 at 1:35












                    Could I ask: How is orientation defined, as you use it?
                    – Yatharth Agarwal
                    Nov 19 at 2:24




                    Could I ask: How is orientation defined, as you use it?
                    – Yatharth Agarwal
                    Nov 19 at 2:24












                    Orientation is the sign of the determinant of the linear transformation (i.e. the basis change matrix): $+$ sign is orientation preserving; $-$ sign is orientation reversing.
                    – Lee Mosher
                    Nov 19 at 2:34




                    Orientation is the sign of the determinant of the linear transformation (i.e. the basis change matrix): $+$ sign is orientation preserving; $-$ sign is orientation reversing.
                    – Lee Mosher
                    Nov 19 at 2:34












                    @LeeMosher That’s a little circular for me, given I’m trying to understand what the sign of the determinant means geometrically. Are there alternative interpretations/definitions?
                    – Yatharth Agarwal
                    yesterday




                    @LeeMosher That’s a little circular for me, given I’m trying to understand what the sign of the determinant means geometrically. Are there alternative interpretations/definitions?
                    – Yatharth Agarwal
                    yesterday












                    Your quest to understand orientation geometrically is a good one, but, my comment was simply an answer to your question in the preceding comment.
                    – Lee Mosher
                    12 hours ago




                    Your quest to understand orientation geometrically is a good one, but, my comment was simply an answer to your question in the preceding comment.
                    – Lee Mosher
                    12 hours ago


















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