Making  sense  of  the  beginning  of  Schoenberg’s  op.  33a.      Schoenberg’s  Klavierstuck  op.  33a  (pp.  62-­‐66  in  your  Roig-­‐Francoli  anthology)  is  a  mature  twelve-­‐tone  work.  By  this  time  in  Schoenberg’s  career,  he  was  finding  ways  to  choose  row  forms  that  would  ensure  even  greater  coherence  and  unity  (coherence  and  unity  were  very  important  to  Schoenberg  and  his  ‘school’).      This  lecture  focuses  on  the  first  five  measures  of  op.  33a,  and  it  uncovers  some  new  answers  to  the  two  questions  that  matter  most:    

• What  factors  influence  row  construction?  • Given  the  48  row-­‐forms  that  any  row  can  provide,  what  factors  influence  row-­‐

form  choice?    Here  is  the  opening  row  of  the  piece,  P10.        10   5   0   11   9   6   1   3   7   8   2   4      It’s  hard  to  identify  P10  at  the  beginning  of  the  piece,  because  groups  of  notes  sound  simultaneously  (and  when  they  do,  it’s  difficult  to  determine  the  order).  Composers  often  let  adjacent  tones  in  a  row  sound  simultaneously,  and  when  they  do,  it  can  present  challenges  to  analysts  until  they  find  a  more  ordered  presentation.  The  integrity  of  the  row  does  become  much  clearer  later  in  the  piece,  though,  so  we  may  rest  assured  that  this  analysis  is  correct  and  turn  our  attention  to  aspects  of  the  composition  that  are  more  immediately  relevant.      Measure  1  consists  of  P10,  broken  down  into  its  discrete  tetrachords  (of  which  there  can  only  be  three).  Measure  2  consists  of  a  similar  breakdown  of  RI3.        9   11   5   6   10     0   7   4   2   1   8   3  

       d.t.11                d.t.2              d.t.3      The  question  arises  of  why  Schoenberg  would  have  chosen  RI3  to  follow  P10.  Comparing  the  two  rows,  there  does  not  seem  to  be  very  much  significant  invariance.  So  what  other  factors  might  have  contributed  to  the  decision  to  follow  the  prime  row  with  an  upside-­‐down  and  backwards  transposition  of  it?  Another  way  to  ask  the  question  is  this:  Given  that  RI3  follows  P10,  how  do  we  make  musical  sense  of  it?  why…  Why…  WHY??    

                                                                                                             1  “d.t.1”  means  ‘Discrete  Tetrachord  1.  

 There  are  several  answers  to  this  question.  It  makes  sense  (as  it  turns  out)  to  address  the  question  in  three  parts.  First,  lets  ask  why  Schoenberg  would  have  chosen  to  follow  the  prime  with  a  backwards  row.  The  answer  has  to  do  with  the  fact  that  he’s  dividing  the  rows  into  discrete  tetrachords.  Across  m.  1,  a  ‘progression’  of  chords  sounds  with  the  following  order  of  set  classes:      10   5   0   11   9   6   1   3   7   8   2   4  

   (0127)          (0258)          (0146)      When  RI3  sounds  in  the  very  next  measure,  we  hear  the  same  progression  of  chords  in  reverse.  This  is  entirely  due  to  the  fact  that  Schoenberg  chose  a  ‘backwards’  kind  of  row.        9   11   5   6   10     0   7   4   2   1   8   3        (0146)            (0258)          (0127)    If  you  play  the  opening  of  this  piece  (or  a  recording),  you’ll  be  able  to  hear  the  progression  of  chord  qualities,  followed  by  the  reverse  of  that  progression.    This  reflects  Schoenberg’s  preoccupation  with  symmetry,  which  was  a  big  concern  of  a  lot  of  composers  during  the  first  half  of  the  20th  century  (also  a  little  before  then,  and  after).  Symmetry  was  a  particularly  important  concern  of  composers  who  chose  to  write  music  outside  the  tonal  system.  This  was  partially  because  symmetry  could  provide  unity  and  balance  to  a  work.      Another  reason  is  that  there  are  many  ways  in  which  a  composer  can  achieve  a  symmetrical  pattern.  This  leads  us  to  our  second  question.  We’ve  just  established  that  a  concern  with  symmetry  factored  into  Schoenberg’s  choice  of  a  backwards  row,  and  that  explains  the  ‘R’  of  RI3.  But  what  about  the  ‘I’  in  RI3?      The  answer  becomes  clear  when  we  investigate  the  voicings  of  the  chords  in  the  first  six  measures.  The  example  below  shows  how  the  tetrachords  with  the  same  prime  forms  (labeled  ‘x,’  ‘y,’  and  ‘z’  in  each  row)  are  voiced  equivalently,  only  upside  down  with  respect  to  each  other.  See  how  the  first  tetrachord  ‘x’  is  voiced  with  ic  1,  5,  and  5,  form  the  bottom  up.  The  intervals  in  tetrachord  ‘x’  of  RI3  are  1,  5,  and  5  from  the  top  down.  The  same  relationship  holds  between  the  ‘y’  tetrachords  and  the  ‘z’  tetrachords,  too.    

   So,  Schoenberg  capitalizes  on  the  ‘I’  aspect  (that  is,  the  upside-­‐down  aspect)  of  this  row-­‐form  choice  by  voicing  the  chords  from  RI3  in  a  perfectly  ‘upside  down’  way  with  respect  to  P10.  This  is  yet  another  kind  of  symmetry.  Schoenberg’s  took  advantage  of  the  ‘R’  aspect  by  creating  a  symmetrical  chord  progression  (x-­‐y-­‐z/z-­‐y-­‐x).  In  a  way,  this  is  horizontal  symmetry.  Schoenberg  took  advantage  of  the  ‘I’  aspect  by  voicing  equivalent  tetrachords  in  the  same  way,  but  upside  down.  This  exploits  what  could  be  regarded  as  vertical  symmetry.      So  if  we  can  answer  “Why  R?”  and  “Why  I?”  (both  answers  being  “  to  achieve  symmetry”),  the  only  question  remains  is  why  Schoenberg  would  choose  the  particular  level  of  transposition  that  he  did.  In  other  words,  “Why  3?”  To  answer  this  question,  we  must  go  a  bit  further  in  the  piece  and  consider  mm.  3-­‐5.    To  analyze  this  passage,  you  must  understand  that  one  row  form  sounds  in  the  upper  register  while  another  sounds  in  the  lower.  (Rows  sound  against  each  other  this  way  over  the  course  of  the  majority  of  the  piece.)  Across  mm.  2-­‐5,  Schoenberg  places  RI3  in  the  upper  register.  9-­‐11-­‐5-­‐6  sounds  across  m.  3,  10-­‐0-­‐7-­‐4  sounds  across  m.  4,  and  2-­‐1-­‐8-­‐3  sounds  across  m.  5  (the  D#  sounds  in  the  left  hand,  but  above  the  right  hand).  In  the  lower  register,  R10  (the  retrograde  of  our  original  P10)  sounds.      Let’s  compare  these  rows.    R10  4   3   8   7   3   1   6   9   11   0   5   10      9   11   5   6   10     0   7   4   2   1   8   3  RI3      If  you  compare  the  pitch-­‐class  content,  and  consider  the  evidence  carefully,  you’ll  notice  that  the  first  six  pitch  classes  of  R10  are  the  same  as  the  last  six  of  RI3  (unordered,  of  course).  Similarly,  the  first  six  pitch  classes  of  RI3  are  the  same  (unordered)  as  the  last  six  

of  R10.  Schoenberg  put  these  rows  against  each  other  precisely  because  they  complement  each  other  in  this  way.  By  the  time  the  first  half  of  each  row  has  sounded,  all  twelve  pitch  classes  have  been  presented.  By  the  time  the  second  half  of  each  row  has  sounded,  all  twelve  pitch  classes  have  occurred  again.  When  rows  sound  against  each  other  like  this,  the  entire  aggregate  of  pitch  classes  sounds  four  times!    

• Once  when  R10  sounds,  • once  when  RI3  sounds,  • once  across  the  first  half  of  each  row,    • once  again  across  the  second  half  of  each  row.  

 Across  op.  33a,  Schoenberg  chooses  pairs  of  rows  and  sets  them  against  each  other  in  different  registers.  Whenever  he  does  this,  the  rows  complement  each  other  so  that  their  halves  have  no  common  tones  (as  discussed  immediately  above).  By  doing  so,  listeners  get  to  hear  twelve-­‐tone  saturation  in  a  number  of  ways.  This  technique  (which  was  used  by  other  composers  such  as  Anton  Webern  and  Milton  Babbitt  as  well)  is  called  Hexachordal  Combinatoriality.      Hexachordal  Combinatoriality  (which  consists  of  twelve  syllables-­‐but  that’s  probably  a  coincidence)  was  the  factor  that  governed  Schoenberg’s  row-­‐form  choice  of  RI3.  It  didn’t  become  apparent  until  he  set  it  against  the  retrograde  of  the  original  row,  but  it  was  a  factor  all  along.      While  Hexachordal  combinatoriality  is  a  factor  that  influenced  Schoenberg’s  row-­‐form  choice,  it  is  also  a  factor  that  can  influence  row  construction.  See  the  appendix  of  set  classes  for  hexachords  in  your  textbook  (p.  264)  and  you’ll  see  what  I  mean.  Hexachords  are  arranged  on  this  page  in  two  ways:  they  are  either  paired  off  or  solitary.      

• Rows  with  solitary  hexachords  contain  those  hexachords  that  can  only  combine  with  other  versions  of  themselves  to  make  a  complete  twelve-­‐tone  row.  (That  is,  if  you  start  a  row  with  (012345),  the  only  hexachord  that  will  generate  a  complete  12-­‐tone  statement  will  be  another  transposition  of  (012345).  

• Rows  with  two  hexachords  contain  pairs  that  create  twelve-­‐tone  rows  together.  (Look  at  the  third  line  down:  The  only  hexachord  that  will  complement  (012347)  to  make  a  complete  12-­‐tone  statement  will  be  (012356).    

The  columns  of  numbers  down  the  left-­‐hand  side  (labeled  P,  R,  I,  RI)  basically  tell  you  this:  If  you  were  to  make  a  twelve-­‐tone  row  out  of  the  hexachord  (or  hexachords)  in  this  row,  for  every  row  form  you  chose,  you’d  have  ‘x’  many  other  rows  that  will  be  hexachordally  combinatorial.      See,  for  example  hexachord  6-­‐7  (012678).  It  can  only  combine  with  another  form  of  itself  to  make  a  complete  12-­‐tone  statement  (that’s  why  there’s  no  other  set  class  in  that  row).  If  you  make  a  row  whose  discrete  hexachords  have  this  prime  form,  then  for  

every  row-­‐form  in  the  matrix  you  choose,  you  will  have  2  other  prime  forms  that  are  combinatorial  with  it,  2  Inverted  forms  that  are  combinatorial  with  it,  2  retrograde  forms  that  are  combinatorial,  and  2  retrograde-­‐inverted  forms  that  are  combinatorial.  If  you  look  about  three  quarters  of  the  way  down  the  page,  you’ll  find  set  class  6-­‐16  (014568).  This  set  also  complements  itself.  For  every  row  form  in  a  matrix  generated  by  complementary  (014568)s,  you’ll  find  no  combinatorial  P-­‐related  rows,  one  complimentary  R-­‐related  row,  one  complimentary  I-­‐related  row,  and  one  complimentary  RI-­‐related  row.    So,  as  you  can  see,  hexachordal  combinatoriality  is  also  a  factor  that  can  influence  row  construction  as  well.  That  is,  composers  can  choose  to  make  rows  out  of  hexachords  that  will  give  them  more  opportunities  for  hexachordal  combinatoriality.  This  idea  of  saturating  listeners’  ears  with  aggregate  completion  answers  both  of  our  ‘big  idea’  questions.      In  conclusion,  we’ve  learned  the  following  from  the  first  five  measures  of  Schoenberg’s  op.  33a.      

• What  factors  influence  row  construction?  Hexachordal  Combinatoriality  • Given  the  48  row-­‐forms  that  any  row  can  provide,  what  factors  influence  row-­‐

form  choice?  Hexachordal  Combinatoriality,  Symmetry